text:
Outlook: PPX Mining Corp. is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Sell
Time series to forecast n: 31 Jan 2023 for (n+4 weeks)
Methodology : Modular Neural Network (News Feed Sentiment Analysis)

## Abstract

PPX Mining Corp. prediction model is evaluated with Modular Neural Network (News Feed Sentiment Analysis) and Sign Test1,2,3,4 and it is concluded that the PPX:TSXV stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period, the dominant strategy among neural network is: Sell

## Key Points

1. How can neural networks improve predictions?
2. What is prediction in deep learning?
3. Can stock prices be predicted?

## PPX:TSXV Target Price Prediction Modeling Methodology

We consider PPX Mining Corp. Decision Process with Modular Neural Network (News Feed Sentiment Analysis) where A is the set of discrete actions of PPX:TSXV stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4

F(Sign Test)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (News Feed Sentiment Analysis)) X S(n):→ (n+4 weeks) $∑ i = 1 n r i$

n:Time series to forecast

p:Price signals of PPX:TSXV stock

j:Nash equilibria (Neural Network)

k:Dominated move

a:Best response for target price

For further technical information as per how our model work we invite you to visit the article below:

How do AC Investment Research machine learning (predictive) algorithms actually work?

## PPX:TSXV Stock Forecast (Buy or Sell) for (n+4 weeks)

Sample Set: Neural Network
Stock/Index: PPX:TSXV PPX Mining Corp.
Time series to forecast n: 31 Jan 2023 for (n+4 weeks)

According to price forecasts for (n+4 weeks) period, the dominant strategy among neural network is: Sell

X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)

Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)

Z axis (Grey to Black): *Technical Analysis%

## IFRS Reconciliation Adjustments for PPX Mining Corp.

1. When a group of items that constitute a net position is designated as a hedged item, an entity shall designate the overall group of items that includes the items that can make up the net position. An entity is not permitted to designate a non-specific abstract amount of a net position. For example, an entity has a group of firm sale commitments in nine months' time for FC100 and a group of firm purchase commitments in 18 months' time for FC120. The entity cannot designate an abstract amount of a net position up to FC20. Instead, it must designate a gross amount of purchases and a gross amount of sales that together give rise to the hedged net position. An entity shall designate gross positions that give rise to the net position so that the entity is able to comply with the requirements for the accounting for qualifying hedging relationships.
2. A net position is eligible for hedge accounting only if an entity hedges on a net basis for risk management purposes. Whether an entity hedges in this way is a matter of fact (not merely of assertion or documentation). Hence, an entity cannot apply hedge accounting on a net basis solely to achieve a particular accounting outcome if that would not reflect its risk management approach. Net position hedging must form part of an established risk management strategy. Normally this would be approved by key management personnel as defined in IAS 24.
3. Fluctuation around a constant hedge ratio (and hence the related hedge ineffectiveness) cannot be reduced by adjusting the hedge ratio in response to each particular outcome. Hence, in such circumstances, the change in the extent of offset is a matter of measuring and recognising hedge ineffectiveness but does not require rebalancing.
4. An entity's business model refers to how an entity manages its financial assets in order to generate cash flows. That is, the entity's business model determines whether cash flows will result from collecting contractual cash flows, selling financial assets or both. Consequently, this assessment is not performed on the basis of scenarios that the entity does not reasonably expect to occur, such as so-called 'worst case' or 'stress case' scenarios. For example, if an entity expects that it will sell a particular portfolio of financial assets only in a stress case scenario, that scenario would not affect the entity's assessment of the business model for those assets if the entity reasonably expects that such a scenario will not occur. If cash flows are realised in a way that is different from the entity's expectations at the date that the entity assessed the business model (for example, if the entity sells more or fewer financial assets than it expected when it classified the assets), that does not give rise to a prior period error in the entity's financial statements (see IAS 8 Accounting Policies, Changes in Accounting Estimates and Errors) nor does it change the classification of the remaining financial assets held in that business model (ie those assets that the entity recognised in prior periods and still holds) as long as the entity considered all relevant information that was available at the time that it made the business model assessment.

*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.

## Conclusions

PPX Mining Corp. is assigned short-term Ba1 & long-term Ba1 estimated rating. PPX Mining Corp. prediction model is evaluated with Modular Neural Network (News Feed Sentiment Analysis) and Sign Test1,2,3,4 and it is concluded that the PPX:TSXV stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period, the dominant strategy among neural network is: Sell

### PPX:TSXV PPX Mining Corp. Financial Analysis*

Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementB3Baa2
Balance SheetB2Caa2
Leverage RatiosBaa2B2
Cash FlowBaa2B2
Rates of Return and ProfitabilityB1Baa2

*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?

### Prediction Confidence Score

Trust metric by Neural Network: 80 out of 100 with 554 signals.

## References

1. M. Benaim, J. Hofbauer, and S. Sorin. Stochastic approximations and differential inclusions, Part II: Appli- cations. Mathematics of Operations Research, 31(4):673–695, 2006
2. Alpaydin E. 2009. Introduction to Machine Learning. Cambridge, MA: MIT Press
3. Athey S, Bayati M, Doudchenko N, Imbens G, Khosravi K. 2017a. Matrix completion methods for causal panel data models. arXiv:1710.10251 [math.ST]
4. Abadir, K. M., K. Hadri E. Tzavalis (1999), "The influence of VAR dimensions on estimator biases," Econometrica, 67, 163–181.
5. Bennett J, Lanning S. 2007. The Netflix prize. In Proceedings of KDD Cup and Workshop 2007, p. 35. New York: ACM
6. S. Bhatnagar. An actor-critic algorithm with function approximation for discounted cost constrained Markov decision processes. Systems & Control Letters, 59(12):760–766, 2010
7. Abadie A, Diamond A, Hainmueller J. 2015. Comparative politics and the synthetic control method. Am. J. Political Sci. 59:495–510
Frequently Asked QuestionsQ: What is the prediction methodology for PPX:TSXV stock?
A: PPX:TSXV stock prediction methodology: We evaluate the prediction models Modular Neural Network (News Feed Sentiment Analysis) and Sign Test
Q: Is PPX:TSXV stock a buy or sell?
A: The dominant strategy among neural network is to Sell PPX:TSXV Stock.
Q: Is PPX Mining Corp. stock a good investment?
A: The consensus rating for PPX Mining Corp. is Sell and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of PPX:TSXV stock?
A: The consensus rating for PPX:TSXV is Sell.
Q: What is the prediction period for PPX:TSXV stock?
A: The prediction period for PPX:TSXV is (n+4 weeks)


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# Limits, Continuity and Differentiability – Method of Rationalization

## Limits, Continuity and Differentiability – Method of Rationalization

Rationalization method is used when, we have radical signs in an expression (like ½, ⅓ etc) and there exists a negative sign between two terms of an algebraic expression.

After rationalization, the terms are factorized which on cancellation gives the required result.

Example: $$\underset{x\to \,\pm \,\infty }{\mathop{\lim }}\,x\left( \sqrt{{{x}^{2}}+k}-x \right),\,k>0$$ is equal to

Solution: $$\underset{x\to \,\pm \,\infty }{\mathop{\lim }}\,x\left( \sqrt{{{x}^{2}}+k}-x \right)\frac{\left( \sqrt{{{x}^{2}}+k}+x \right)}{\sqrt{{{x}^{2}}+k}+x}$$

$$=\underset{x\to \,\pm \,\infty }{\mathop{\lim }}\,\frac{x\left( {{x}^{2}}+k-{{x}^{2}} \right)}{\left( \sqrt{{{x}^{2}}+k}+x \right)}=\underset{x\to \,\pm \,\infty }{\mathop{\lim }}\,\frac{xk}{\left( |x|\sqrt{\left( 1+\frac{k}{{{x}^{2}}} \right)} \right)+x}$$

Here, we have to consider two cases.

i) When x → ∞; |x| = x

Then, the given limit $$=\underset{x\to \infty }{\mathop{\lim }}\,\frac{xk}{x\sqrt{\left( 1+\frac{k}{{{x}^{2}}} \right)}+x}$$

$$=\underset{x\to \infty }{\mathop{\lim }}\,\frac{xk}{x\left( \sqrt{\left( 1+\frac{k}{{{x}^{2}}} \right)}+1 \right)}=\frac{k}{2}$$

ii) When x → -∞; |x| = -x

Then, we have$$\underset{x\to -\infty }{\mathop{\lim }}\,\frac{xk}{-x\sqrt{\left( 1+\frac{k}{{{x}^{2}}} \right)}+x}$$

$$\underset{x\to -\infty }{\mathop{\lim }}\,\frac{xk}{x\left( -\sqrt{\left( 1+\frac{k}{{{x}^{2}}} \right)}+1 \right)}=\frac{k}{-{{1}^{-}}+1}=\frac{k}{{{0}^{-}}}\to -\infty$$.

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= Find its approximate solution using Euler method. Once again, we can see why we needed to require $$x > 0$$. {\displaystyle t_{0}} t This shows that for small Now, one step of the Euler method from is evaluated at the end point of the step, instead of the starting point. Euler’s theorem states that if a function f(a i, i = 1,2, …) is homogeneous to degree “k”, then such a function can be written in terms of its partial derivatives, as follows: k λ k − 1 f (a i) = ∑ i a i (∂ f (a i) ∂ (λ a i)) | λ x This equation is not rendering properly due to an incompatible browser. , = = 7. y [ These types of differential equations are called Euler Equations. It is the difference between the numerical solution after one step, $${\displaystyle y_{1}}$$, and the exact solution at time $${\displaystyle t_{1}=t_{0}+h}$$. ) In other words, since $$\eta>0$$ we can use the work above to get solutions to this differential equation. Given a differential equation dy/dx = f(x, y) with initial condition y(x0) = y0. With this transformation the differential equation becomes. is Lipschitz continuous in its second argument, then the global truncation error (GTE) is bounded by, where , [4], we would like to use the Euler method to approximate The global truncation error is the cumulative effect of the local truncation errors committed in each step. ) is known (see the picture on top right). A This conversion can be done in two ways. For the exact solution, we use the Taylor expansion mentioned in the section Derivation above: The local truncation error (LTE) introduced by the Euler method is given by the difference between these equations: This result is valid if = y It is the most basic explicit method for numerical integration of ordinary differential equations and is the simplest Runge–Kutta method. However, this is now a solution for any interval that doesn’t contain $$x = 0$$. . t For this reason, the Euler method is said to be a first-order method, while the midpoint method is second order. For a class of nonlinear impulsive fractional differential equations, we first transform them into equivalent integral equations, and then the implicit Euler method is adapted for solving the problem. Most of the effect of rounding error can be easily avoided if compensated summation is used in the formula for the Euler method.[20]. and can be handled by Euler's method or, in fact, by any other scheme for first-order systems. Derivations. July 2020 ; Authors: Zimo Hao. = h {\displaystyle A_{1}.} In step n of the Euler method, the rounding error is roughly of the magnitude εyn where ε is the machine epsilon. {\displaystyle (0,1)} = . It is customary to classify them into ODEs and PDEs.. We construct the general solution by using the trial power function $$y = {x^k}.$$ Substitute the derivatives of this function into the differential equation: A The general nonhomogeneous differential equation is given by x^2(d^2y)/(dx^2)+alphax(dy)/(dx)+betay=S(x), (1) and the homogeneous equation is x^2y^('')+alphaxy^'+betay=0 (2) y^('')+alpha/xy^'+beta/(x^2)y=0. 1 y That is, we can't solve it using the techniques we have met in this chapter (separation of variables, integrable combinations, or using an integrating factor), or other similar means. f {\displaystyle f} h To find the constants we differentiate and plug in the initial conditions as we did back in the second order differential equations chapter. {\displaystyle y_{1}} y(0) = 1 and we are trying to evaluate this differential equation at y = 1. n n is outside the region. {\displaystyle y_{3}} . will be close to the curve. Note that while this does not involve a series solution it is included in the series solution chapter because it illustrates how to get a solution to at least one type of differential equation at a singular point. Recall from the previous section that a point is an ordinary point if the quotients. Euler’s Method, is just another technique used to analyze a Differential Equation, which uses the idea of local linearity or linear approximation, where we use small tangent lines over a short distance to approximate the solution to an initial-value problem. t 2.3 {\displaystyle f(t_{0},y_{0})} h 5.2. {\displaystyle \mathbf {z} (t)} t y t , then the numerical solution is unstable if the product {\displaystyle (t-t_{0})/h} y t − ( = = ξ Here, a differential equation can be thought of as a formula by which the slope of the tangent line to the curve can be computed at any point on the curve, once the position of that point has been calculated. 4 h What is Euler’s Method?The Euler’s method is a first-order numerical procedure for solving ordinary differential equations (ODE) with a given initial value. Viewed 1k times 10. for In this section we want to look for solutions to. Assuming that the rounding errors are all of approximately the same size, the combined rounding error in N steps is roughly Nεy0 if all errors points in the same direction. It can be reduced to the linear homogeneous differential equation with constant coefficients. We terminatethis pr… {\displaystyle 1/h} Since the number of steps is inversely proportional to the step size h, the total rounding error is proportional to ε / h. In reality, however, it is extremely unlikely that all rounding errors point in the same direction. So, we get the roots from the identical quadratic in this case. Now, we assumed that $$x>0$$ and so this will only be zero if. Output of this is program is solution for dy/dx = x + y with initial condition y = 1 for x = 0 i.e. E on E ano ahni, itu ahni, auar era, shnil andaliya, hairya hah E olue , certain kind of uncertainty. t y The error recorded in the last column of the table is the difference between the exact solution at As a result, we need to resort to using numerical methods for solving such DEs. ∈ t h For many of the differential equations we need to solve in the real world, there is no "nice" algebraic solution. , then the numerical solution is qualitatively wrong: It oscillates and grows (see the figure). + {\displaystyle k} Wuhan University; Michael Röckner. {\displaystyle y_{i}} We chop this interval into small subdivisions of lengthh. Euler Method Online Calculator. Differential Equations play a major role in most of the science applications. x. in a first-year calculus context, and the MacLaurin series for. 2. Key–Words: Fractional differential equations, Initial value problem, Solution, Existence, Eulers method 1 Introduction With the rapid development of high-tech, the frac-tional calculus gets involved in more and more ar-eas, especially in control theoryviscoelastic theory-electronic chemicalsfractal theory and so on. y A {\displaystyle t_{n}=t_{0}+nh} Find its approximate solution using Euler method. / {\displaystyle \xi \in [t_{0},t_{0}+h]} Differential Equations Notes PDF. {\displaystyle A_{0}} 0 The first derivation is based on power series, where the exponential, sine and cosine functions are expanded as power series to conclude that the formula indeed holds.. , which we take equal to one here: Since the step size is the change in More accurate second-order Runge-Kutta methods have the form k1= Dxf(xn,y), k2= Dxf(x +aDx,y +bk1), yn+1= yn+ ak1+bk2. e Euler Equations; In the next three sections we’ll continue to study equations of the form $\label{eq:7.4.1} P_0(x)y''+P_1(x)y'+P_2(x)y=0$ where $$P_0$$, $$P_1$$, and $$P_2$$ are polynomials, but the emphasis will be different from that of Sections 7.2 and 7.3, where we obtained solutions of Equation \ref{eq:7.4.1} near an ordinary point $$x_0$$ in the form of power series in $$x-x_0$$. For this reason, the Euler method is said to be first order. {\displaystyle t} {\displaystyle f(t,y)=y} . "Eulers theorem for homogeneous functions". If we didn’t we’d have all sorts of problems with that logarithm. We can make one more generalization before working one more example. Our results are stronger because they work in any dimension and yield bounded velocity and pressure. {\displaystyle h=1} + Then, using the initial condition as our starting point, we generatethe rest of the solution by using the iterative formulas: xn+1 = xn + h yn+1 = yn + hf(xn, yn) to find the coordinates of the points in our numerical solution. To deal with this we need to use the variable transformation. {\displaystyle z_{1}(t)=y(t),z_{2}(t)=y'(t),\ldots ,z_{N}(t)=y^{(N-1)}(t)} This large number of steps entails a high computational cost. , and the exact solution at time (Here y = 1 i.e. ( The top row corresponds to the example in the previous section, and the second row is illustrated in the figure. A simple modification of the Euler method which eliminates the stability problems noted in the previous section is the backward Euler method: This differs from the (standard, or forward) Euler method in that the function y + For this reason, people usually employ alternative, higher-order methods such as Runge–Kutta methods or linear multistep methods, especially if a high accuracy is desired.[6]. Conventional theory of differential equation fails to handle this kind of vagueness. Another test example is the initial value problem y˙ = λ(y−sin(t))+cost, y(π/4) = 1/ √ 2, where λis a parameter. Let’s start off by assuming that $$x>0$$ (the reason for this will be apparent after we work the first example) and that all solutions are of the form. h working rule of eulers theorem. ( and we can ask for solutions in any interval not containing $$x = {x_0}$$. y Much like the familiar oceanic waves, waves described by the Euler Equations 'break' and so-called shock waves are formed; this is a nonlinear effect and represents the solution becoming multi-valued. y . {\displaystyle h=1} Xicheng Zhang. This is what it means to be unstable. Although the approximation of the Euler method was not very precise in this specific case, particularly due to a large value step size A. {\displaystyle y'=f(t,y)} h Online tool to solve ordinary differential equations with initial conditions (x0, y0) and calculation point (xn) using Euler's method. Other modifications of the Euler method that help with stability yield the exponential Euler method or the semi-implicit Euler method. A slightly different formulation for the local truncation error can be obtained by using the Lagrange form for the remainder term in Taylor's theorem. y Euler's Method after the famous Leonhard Euler. {\displaystyle h} y [15], The precise form of this bound is of little practical importance, as in most cases the bound vastly overestimates the actual error committed by the Euler method. illustrated on the right. The numerical solution is given by y In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Consider the 1st-order Cauchy-Euler equation, in a multivariate extension: $$a_1\mathbf x'\cdot \nabla f(\mathbf x) + a_0f(\mathbf x) = 0 \tag{3}$$ 0 Of course, in practice we wouldn’t use Euler’s Method on these kinds of differential equations, but by using easily solvable differential equations we will be able to check the accuracy of the method. f The concept is similar to the numerical approaches we saw in an earlier integration chapter (Trapezoidal Rule, Simpson's Rule and Riemann Su… : is smaller. If . and apply the fundamental theorem of calculus to get: Now approximate the integral by the left-hand rectangle method (with only one rectangle): Combining both equations, one finds again the Euler method. {\displaystyle y_{n}\approx y(t_{n})} As suggested in the introduction, the Euler method is more accurate if the step size t 1 We should now talk about how to deal with $$x < 0$$ since that is a possibility on occasion. ( to 0 t 1 k We’ll get two solutions that will form a fundamental set of solutions (we’ll leave it to you to check this) and so our general solution will be. y = {\displaystyle y(t)=e^{t}} This is a fourth-order homogeneous Euler equation. Was Euler's theorem in differential geometry motivated by matrices and eigenvalues? z + 0 We first need to find the roots to $$\eqref{eq:eq3}$$. . and so the general solution in this case is. + ( We’ll get two solutions that will form a fundamental set of solutions (we’ll leave it to you to check this) and so our general solution will be,With the solution to this example we can now see why we required x>0x>0. y t The Euler method gives an approximation for the solution of the differential equation: $\frac{dy}{dt} = f(t,y) \tag{6}$ with the initial condition: $y(t_0) = y_0 \tag{7}$ where t is continuous in the interval [a, b]. N h {\displaystyle t_{n+1}=t_{n}+h} In this simple differential equation, the function 3 3 n 7 $\begingroup$ I am teaching a class on elementary differential geometry and I would like to know, for myself and for my students, something more about the history of Euler Theorem and Euler equation: the curvature of a … Help to clarify proof of Euler's Theorem on homogenous equations. We show that any such flow is a shear flow, that is, it is parallel to some constant vector. If instead it is assumed that the rounding errors are independent random variables, then the expected total rounding error is proportional to {\displaystyle y(4)=e^{4}\approx 54.598} ) {\displaystyle f} t z. since this result requires complex analysis. This value is then added to the initial The above steps should be repeated to find h 0 If the solution ( {\displaystyle y'=f(t,y)} y f (x, y), y(0) y 0 dx dy = = (1) So only first order ordinary differential equations can be solved by using Euler’s method. {\displaystyle M} Show Instructions. Now, as we’ve done every other time we’ve seen solutions like this we can take the real part and the imaginary part and use those for our two solutions. The other possibility is to use more past values, as illustrated by the two-step Adams–Bashforth method: This leads to the family of linear multistep methods. And not only actually is this one a good way of approximating what the solution to this or any differential equation is, but actually for this differential equation in particular you can actually even use this … = In this case it can be shown that the second solution will be. {\displaystyle h} ( A more general form of an Euler Equation is. y′ + 4 x y = x3y2. This is true in general, also for other equations; see the section Global truncation error for more details. The exact solution is on both sides, so when applying the backward Euler method we have to solve an equation. 4 . Eulers theorem in hindi. . t Finally, one can integrate the differential equation from We show a coincidence of index of rigidity of differential equations with irregular singularities on a compact Riemann surface and Euler characteristic of the associated spectral curves which are recently called irregular spectral curves. {\displaystyle A_{1}} This equation is a quadratic in $$r$$ and so we will have three cases to look at : Real, Distinct Roots, Double Roots, and Complex Roots. Again, this yields the Euler method. {\displaystyle y_{n+1}} can be computed, and so, the tangent line. » Differential Equations » 11. 0 {\displaystyle t_{n}} h divided by the change in value to obtain the next value to be used for computations. Indeed, it follows from the equation y : The differential equation states that t . [16] What is important is that it shows that the global truncation error is (approximately) proportional to [22], For integrating with respect to the Euler characteristic, see, % equal to: t0 + h*n, with n the number of steps, % i yi ti f(yi,ti), % 0 +1.00 +0.00 +1.00, % 1 +2.00 +1.00 +2.00, % 2 +4.00 +2.00 +4.00, % 3 +8.00 +3.00 +8.00, % 4 +16.00 +4.00 +16.00, % NOTE: Code also outputs a comparison plot. ) y The second derivation of Euler’s formula is based on calculus, in which both sides of the equation are treated as functions and differentiated accordingly. t t has a bounded third derivative.[10]. ) y e around , and the error committed in each step is proportional to In this scheme, since, the starting point of each sub-interval is used to find the slope of the solution curve, the solution would be correct only if the function is linear. {\displaystyle A_{0}A_{1}A_{2}A_{3}\dots } f The Euler method is explicit, i.e. / Differential equation Calculates the solution y=f(x) of the ordinary differential equation y'=F(x,y) using Euler's method. The Euler method for solving the differential equation dy/dx = f(x,y) can be rewritten in the form k1= Dxf(xn,y), yn+1= yn+k1, and is called a ﬁrst-order Runge-Kutta method. n has a bounded second derivative and ) Whenever an A and B molecule bump into each other the B turns into an A: A + B ! Consider the problem of calculating the shape of an unknown curve which starts at a given point and satisfies a given differential equation. above can be used. The local truncation error of the Euler method is the error made in a single step. Thus, it is to be expected that the global truncation error will be proportional to The value of {\displaystyle h} {\displaystyle y} , The MacLaurin series: Warning 1 You might be wondering what is suppose to mean: how can we differentiate with respect to a derivative? 0 n 1 , In this case we’ll be assuming that our roots are of the form. Differential Equations + Euler + Phasors Christopher Rose ABSTRACT You have a network of resistors, capacitors and inductors. 2 {\displaystyle y} to treat the equation. The convergence analysis of the method shows that the method is convergent of the first order. h , after however many steps the methods needs to take to reach that time from the initial time. We give a reformulation of the Euler equations as a differential inclusion, and in this way we obtain transparent proofs of several celebrated results of V. Scheffer and A. Shnirelman concerning the non-uniqueness of weak solutions and the existence of energy-decreasing solutions. i = 1 16 [5], so first we must compute ) ty′ + 2y = t2 − t + 1. on the given interval and = {\displaystyle t_{n}} k f For my math investigation project, I was trying to predict the trajectory of an object in a projectile motion with significant air resistance by using the Euler's Method. Date: 1st Jan 2021. ) ( Euler's Method. and {\displaystyle \Delta y/\Delta t} The first fundamental theorem of calculus states that if is a continuous function in the interval [a,b], and is the antiderivative of , then. t A Given a differential equation dy/dx = f(x, y) with initial condition y(x0) = y0. The numerical results verify the correctness of the theoretical results. {\displaystyle L} Conjectures. {\displaystyle f} f In mathematics and computational science, the Euler method (also called forward Euler method) is a first-order numerical procedure for solving ordinary differential equations (ODEs) with a given initial value. The local truncation error of the Euler method is the error made in a single step. − ) This region is called the (linear) stability region. N

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What is the mass of oxalic acid, ${H_2}{C_2}{O_4}$ , which can be oxidized by $C{O_2}$ by $100mL$ of $MnO_4^ -$ solution, $10mL$ of which is capable of oxidizing $50mL$ of $1.00N$ ${I^ - }$ to ${I_2}$ ?A. $2.25g$B. $52.2g$C. $25.2g$D. $22.5g$

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Hint: We need to realize that oxalic corrosive is having the equation ${H_2}{C_2}{O_4}$ . It is likewise called crab corrosive. It is really a white glasslike strong which is by and large found to get broken up in arrangement and colourless. The equation used to ascertain the load for oxalic corrosive must be given underneath,
${N_1}{V_1} = \dfrac{{Weight \times Valencyfactor}}{{Molarmass}} \times 1000$
Here,
${N_1}$ = Normality of $MnO_4^ -$
${V_1}$ = Volume of $MnO_4^ -$

We need to know, the arrangement that is the $100mL$ of $MnO_4^ -$ , at that point it is inquired as to whether we add $10mL$ in $50mL$ of $10N$ ${I^ - }$ , at that point it will be completely changed over into , so we can compose its condition,
$M{n^{ + 7}} + 5{e^ - } \to M{n^{ + 2}}$
Then,
${I^ - } + {e^ - } \to {I_2}$
We can write the expression as,
${N_1}{V_1} = {N_2}{V_2}$
Here,
${N_1}$ = Normality of $MnO_4^ -$
${V_1}$ = Volume of $MnO_4^ -$
${N_2}$ = Normality of ${I^ - }$
${V_2}$ = Volume of ${I^ - }$
To calculate the normality ${N_1}$ ,
Applying given values in the above expression,
${N_1} \times 10 = 50 \times 10$
Then,
${N_1} = 5$
Now, we have to calculate mass of the ${H_2}{C_2}{O_4}$ by the use of the value of molar mass of ${H_2}{C_2}{O_4}$ .The molar mass of ${H_2}{C_2}{O_4}$ = $90g/mol$ .
Where, we can see that carbon in oxalic corrosive in $+ 3$ state, and it is evolving to, so the valency of this is $2$ , and reciprocals are something similar.
${N_1}{V_1} = \dfrac{{Weight \times Valencyfactor}}{{Molarmass}} \times 1000$
Applying given values in the above equation,
$5 \times 10 = \dfrac{{w \times 2}}{{90}} \times 100$
Then, to find out the weight,
$w = \dfrac{{5 \times 10 \times 90}}{{2 \times 100}} = \dfrac{{4500}}{{200}} = 22.5$
Therefore,
The weight is $22.5g$

Hence, the option (D) is correct.

Note:
We need to establish ordinariness too. In this way, we ought not to get confused regarding ordinariness and molarity. At the point when ordinariness is the quantity of the same solute disintegrated per liter of arrangement. $N$ is represented as normality, and $M$ is represented as molarity.


text:
# I Riemann Tensor knowing Christoffel symbols (check my result)

#### Confused Physicist

I need to find all the non-zero components of the Riemann Tensor in a two-dimensional geometry knowing that the only two non-zero components of the Christoffel symbols are:

$$\Gamma^x_{xx}=\frac{1}{x}$$ and $$\Gamma^y_{yy}=\frac{2}{y}$$

knowing that: $$R^\alpha_{\beta\gamma\delta}=\partial_\gamma \Gamma^\alpha_{\delta\beta}-\partial_\delta \Gamma^\alpha_{\gamma\beta}+\Gamma^\epsilon_{\delta\beta}\Gamma^\alpha_{\gamma\epsilon}-\Gamma^\epsilon_{\gamma\beta}\Gamma^\alpha_{\delta\epsilon}$$

The result I have obtained is that all the components of the Riemann curvature tensor are zero. Is this correct? If it is, what does it mean that all the components are zero?

MY PROCEDURE HAS BEEN:

the only plausible non-zero components of the Riemann curvature tensor are:
$$R^\alpha_{\beta\gamma\delta}=\partial_\gamma \Gamma^\alpha_{\delta\beta}-\partial_\delta \Gamma^\alpha_{\gamma\beta}+\Gamma^x_{\delta\beta}\Gamma^\alpha_{\gamma x}+\Gamma^y_{\delta\beta}\Gamma^\alpha_{\delta y}-\Gamma^x_{\gamma\beta}\Gamma^\alpha_{\delta x}-\Gamma^y_{\gamma\beta}\Gamma^\alpha_{\delta y}$$

$$\alpha=x\quad:\quad R^x_{\beta\gamma\delta}=\partial_\gamma \Gamma^x_{\delta\beta}-\partial_\delta \Gamma^x_{\gamma\beta}+\Gamma^x_{\delta\beta}\Gamma^x_{\gamma x}+\Gamma^y_{\delta\beta}\Gamma^x_{\delta y}-\Gamma^x_{\gamma\beta}\Gamma^x_{\delta x}-\Gamma^y_{\gamma\beta}\Gamma^x_{\delta y}$$

$$\text{ }\quad\quad \Longrightarrow \quad R^x_{xxx}=\partial_x \Gamma^x_{xx}-\partial_x \Gamma^x_{xx}+\Gamma^x_{xx}\Gamma^x_{xx}+\Gamma^y_{xx}\Gamma^x_{xy}-\Gamma^x_{xx}\Gamma^x_{xx}-\Gamma^y_{xx}\Gamma^x_{xy}=0$$

$$\alpha=y\quad:\quad R^y_{\beta\gamma\delta}=\partial_\gamma \Gamma^y_{\delta\beta}-\partial_\delta \Gamma^y_{\gamma\beta}+\Gamma^x_{\delta\beta}\Gamma^y_{\gamma x}+\Gamma^y_{\delta\beta}\Gamma^y_{\delta y}-\Gamma^x_{\gamma\beta}\Gamma^y_{\delta x}-\Gamma^y_{\gamma\beta}\Gamma^y_{\delta y}$$

$$\text{ }\quad\quad \Longrightarrow \quad R^y_{yyy}=\partial_y \Gamma^y_{yy}-\partial_y \Gamma^y_{yy}+\Gamma^y_{yy}\Gamma^y_{yx}+\Gamma^y_{yy}\Gamma^y_{yy}-\Gamma^y_{yy}\Gamma^y_{yx}-\Gamma^y_{yy}\Gamma^x_{yy}=0$$

Therefore, all the components of the Riemann Tensor are zero.

Thanks!!!

Last edited:
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#### Orodruin

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Your conclusion is correct and your argument holds. However, your argument would be made more concise if, instead of checking the cases for $\alpha$, you used the antisymmetry of the curvature in the two last indices.

If it is, what does it mean that all the components are zero?
Are you familiar with the geometrical interpretation of the curvature tensor?

#### Confused Physicist

Are you familiar with the geometrical interpretation of the curvature tensor?
No, not really. Could you illustrate it for me? Thank you!!

#### Orodruin

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How was the curvature tensor introduced to you?

#### Confused Physicist

How was the curvature tensor introduced to you?
Purely mathematical. Two years ago I took a course on differential geometry. It wasn't until this year I started studying General Relativity, but I'm learning it on my own.

#### Orodruin

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Purely mathematical.
This does not help. Geometry (also differential geometry) is a mathematics subfield so I would think even a ”pure maths” intro would mention the geometrical interpretation. The question I am asking is how it was presented to you.

#### Confused Physicist

This does not help. Geometry (also differential geometry) is a mathematics subfield so I would think even a ”pure maths” intro would mention the geometrical interpretation. The question I am asking is how it was presented to you.
It was presented to me by its deffinition with Christoffel symbols. I was never explained the physical meaning behind it (geometrical meaning) to help me imagine it.

#### Orodruin

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Now that is just bad teaching. The curvature tensor is related to the change of a vector that is parallel transported around a loop. This should be discussed in any introductory text on differential geometry.

#### haushofer

See e.g. chapter 3 of Carroll's notes,

https://arxiv.org/abs/gr-qc/9712019

page 74 (eqn. 3.63) and 75 (eqn.3.65). In short, the Riemann tensor tells you how the orientation of a vector parallelly transported around an infinitesimal loop is changed. This is expressed as the commutator of the two corresponding covariant derivatives. To go with Orodruin, it's hard for me to believe you've never seen this definition; it's like teaching calculus and integration without mentioning the Riemann sum or definition of a derivative.

#### Orodruin

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This is expressed as the commutator of the two corresponding covariant derivatives
This is only true for vector fields that commute (such as the holonomic basis vectors). If the fields do not commute the loop does not close and you need to add a contribution from going along the curve that closes the loop, proportional to the commutator of the fields. The full expression is $R(X,Y)Z = (\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]})Z$.

#### haushofer

Yes, I was a bit too implicit.

#### stevendaryl

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This is only true for vector fields that commute (such as the holonomic basis vectors). If the fields do not commute the loop does not close and you need to add a contribution from going along the curve that closes the loop, proportional to the commutator of the fields. The full expression is $R(X,Y)Z = (\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]})Z$.
@haushofer was talking about parallel-transporting vectors, not vector fields. Can we equally well take parallel transport of vectors to be fundamental, and define covariant derivatives of vector fields in terms of parallel transport, or take covariant derivatives of vector fields as fundamental and define parallel transport of vectors in terms of that?

#### Orodruin

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@haushofer was talking about parallel-transporting vectors, not vector fields
This is a bit of a funny point. In order for $\nabla_X Z$ and $[X,Y]$ to be defined, $X, Y, Z$ need to be vector fields. However, the properties of the objects actually result in the curvature depending only on the vectors at the point in question (if not, it would not be a tensor). Either way, the parallel transport is of a vector $Z$ around a loop spanned by $X$ and $Y$.

Can we equally well take parallel transport of vectors to be fundamental, and define covariant derivatives of vector fields in terms of parallel transport, or take covariant derivatives of vector fields as fundamental and define parallel transport of vectors in terms of that?
Are you asking if the parallel transport equations have a one-to-one correspondence with the connection? The answer to that question is yes. If you know how vectors behave under (all possible) parallel transports, then you know what the connection is, since you will know (in particular) $\nabla_{\dot\gamma}X$ for all curves $\gamma$ and all fields $X$ along $\gamma$.

#### stevendaryl

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This is a bit of a funny point. In order for $\nabla_X Z$ and $[X,Y]$ to be defined, $X, Y, Z$ need to be vector fields.
It seems to me that $\nabla_X Z$ only requires that $Z$ be a vector field, not $X$.

#### Orodruin

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It seems to me that $\nabla_X Z$ only requires that $Z$ be a vector field, not $X$.
Yes, unfortunate formulation, but in the definition of the curvature, all fields have derivatives acting on them and therefore need to be fields. The point is it does not matter what the fields are as long as they agree with the corresponding vectors at the relevant point.

#### stevendaryl

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Yes, unfortunate formulation, but in the definition of the curvature, all fields have derivatives acting on them and therefore need to be fields. The point is it does not matter what the fields are as long as they agree with the corresponding vectors at the relevant point.
I would say that the definition of curvature in terms of covariant derivatives requires all three arguments to be vector fields. But if you define curvature in terms of parallel transport, they don't need to be vector fields, it seems to me.

#### Orodruin

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I would say that the definition of curvature in terms of covariant derivatives requires all three arguments to be vector fields. But if you define curvature in terms of parallel transport, they don't need to be vector fields, it seems to me.
Well, the point is that the curvature tensor is a tensor. It does not depend on the vectors it take being fields. You can extend the vectors $XYZ$ to arbitrary vector fields at will and $R(X,Y)Z$ will still be the same. As such, it is clear that the definition of the curvature tensor only requires vectors in $T_p M$, not sections of $TM$. However, expressing the tensor in terms of the commutator of the covariant derivatives, as done in #10, does require the extension of $XYZ$ to sections (even if the actual extensions are irrelevant). Either way, the parallel transport - connection link is there.

"Riemann Tensor knowing Christoffel symbols (check my result)"

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text:
# Period ofWeeks() method in Java

Java 8Object Oriented ProgrammingProgramming

The Period can be obtained with the given number of weeks using the ofWeeks() method in the Period class in Java. This method requires a single parameter i.e. the number of weeks and it returns the Period object with the given number of weeks.

A program that demonstrates this is given as follows −

## Example

Live Demo

import java.time.Period;
public class Demo {
public static void main(String[] args) {
int weeks = 7;
Period p = Period.ofWeeks(weeks);
System.out.println("The Period is: " + p);
}
}

## Output

The Period is: P49D

Now let us understand the above program.

The Period is obtained with the given number of weeks using the ofWeeks() method and then it is printed. A code snippet that demonstrates this is as follows −

int weeks = 7;
Period p = Period.ofWeeks(weeks);
System.out.println("The Period is: " + p);
Updated on 30-Jul-2019 22:30:25


text:
## Off-On Fourier Series

I have been fascinated lately with the concept of frequency spectrum and the idea that all periodic signals can be approximated by an infinite sum of sinusoidal functions. There are many introductory videos on this subject, usually titled as introductions to the Fourier transform.

As far as the actual math involved, this YouTube* video was very helpful:
Compute Fourier Series Representation of a Function

He converts an off-on type of function to a fourier series. After the integration, we get this:

I translated that into some plots in Racket, to give the visual idea. Say we only add in a single sinusoid:

Then, another:

And a few more:

And a lot more:

And finally, hundreds of them:

It cannot quite perfectly represent the function, because the Fourier series adds an extra point in between the switch from off to on (and back), whereas the original just jumps from 0 to 1 (and back).

Here is the Racket code for those interested (I did not bother to optimize):

#lang racket

(require plot)

(define (pulse x l)
(letrec ([pulse_
(lambda (acc n)
(if (> n l) acc
(pulse_
(+ acc
(/ (* 2
(sin (* (+ (* 2 n) 1) pi x)))
(* (+ (* 2 n) 1) pi)))
(+ n 1))))])
(pulse_ 0.5 0)))

(define (pulseplot l)
(plot
(function (lambda (x) (pulse x l)) -1 3)))

## More Emacs Calculator Functionality

The first video demonstrates algebraic formulas and live evaluation in Emacs Calculator:

The second video covers two subjects: (1) mapping functions over vectors, and (2) using emacs to display algebraic formulas as math LaTeX (to paste into a WordPress post, for example). Please forgive the improper pronunciation of “LaTeX”, which I remembered afterwards.

## Emacs Calc: Angle Between Vectors

In my geometry studies, I learned that one can get the angle between two vectors with this formula:

$cos \theta = \frac{V_1 \cdot V_2}{| V_1 | | V_2 |}$

I.e., the cosine of the angle equals the dot product of the two vectors over the product of their magnitudes.

This is definitely doable in Emacs Calc, since we have a dot product function, called inner-product (press ‘x inner product’), But doing the angle formula involves a lot of steps, with either stack rotation or storing the vectors in variables. So I wanted to get the angle formula stored as a calc formula. Unfortunately, inner-product itself is only an interactive function, so this was problematic. However, inner-product actual calls another function, inner. So, this formula is possible:

arccos(inner(mul, add, v1, v2) / (abs(v1) abs(v2)))

How do you store this formula in Emacs? I could walk you through the steps described in section 18.4 of the Emacs Calc info manual, but the end result is that this code is stored in your ~/.emacs.d/calc.el:

(put 'calc-define 'calc-vectorsangle '(progn
(defun calc-vectorsangle nil (interactive) (calc-wrapper (calc-enter-result 2 "vect" (cons (quote calcFunc-vectorsangle) (calc-top-list-n 2)))))
(put 'calc-vectorsangle 'calc-user-defn 't)
(defun calcFunc-vectorsangle (v1 v2) (math-normalize (list (quote
calcFunc-arccos) (list (quote /) (list (quote calcFunc-inner) (quote
(var mul var-mul)) (quote (var add var-add)) v1 v2) (list (quote *)
(list (quote calcFunc-abs) v1) (list (quote calcFunc-abs) v2))))))
(put 'calcFunc-vectorsangle 'calc-user-defn '(calcFunc-arccos (/
(var v2 var-v2)) (* (calcFunc-abs (var v1 var-v1)) (calcFunc-abs (var
v2 var-v2))))))
(define-key calc-mode-map "zA" 'calc-vectorsangle)
))

Then when you start calc, you can put two vectors on the stack, and enter command ‘x vectorsangle’ or ‘z A’ for short:

This will work for vectors with as many more coordinates as you want, so long as there are the same number in each vector.

## Trigonometry Problem without Trigonometry

Here is a classic trigonometry problem:

With the satellite dish pointed up 13 degrees above the ground, how far up will the Line of Sight (LOS) be after passing over 100 meters of ground? (To save a few sentences in our post, we will ignore the fact that the satellite dish will usually start a few meters above the ground.)

This is simple using any calculator with the trigonometry function tangent:

tan(ϕ) = opp/adj, so height = 100 * tan 13° (after 13° is converted to radians).

If you didn’t want to use a trigonometric function, however, what could you do? Another approach is complex numbers, and it is pretty simple. You simply need to rotate ϕ degrees up the unit circle (a basic operation with complex numbers), get the real and imaginary parts of that complex number, and use that proportion to figure out the opposite side of your problem triangle.

So, after raising e^i by 13 degrees (converted to radians) multiply 100 x (imag Z / real Z). In programming code, it looks like this:

#lang racket

(define tau (* 2 pi))

(define rfact (/ tau 360))

(define (rise gnd deg)
(/ (* gnd (imag-part Z)) (real-part Z))))

> (rise 100 13)
23.08681911255631

It gives geometrically sane results also if you use negative degrees or negative distances:

> (rise 100 -13)
-23.08681911255631
> (rise -100 -13)
23.08681911255631
> (rise -100 13)
-23.08681911255631

What if you don’t have or want to use the exp function (to raise e to some value). You instead start with an approximation of e^i, which happens to be

0.5403023058681398+0.8414709848078965i

and then just raise that to ϕ.

Now, why would you want to use the complex number approach rather than the trig function? Well, to be honest I couldn’t think of a really compelling reason off the top of my head. An easy one would be if you were already using complex numbers for other reasons. Another would be if you wanted to use purely algebraic operations in your programming, say, on a computer board that did not have trigonometric functions built in. I suspect that the complex number approach is ultimately simpler and more efficient on the computational level, but since trigonometric functions are usually approximated with tables and with an algebraic exponential function, I couldn’t really say for sure off the top of my head.

In any case, it seems cool that complex numbers allow you to do an angle and distance trigonometry problem without trigonometry.

## Rotary Harmonograph with Dampening

To get some more artistic variety in the simulated rotary harmonograph output, we introduce a dampening, i.e., causing the pen and table to run down as though from friction. Also, we can nudge our simple ratio slightly, to add a small element of discord or imperfection.

(define (rot ang/tunit t phase dampfact)
(+ phase (* ang/tunit t)))))


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# Bounciness in Rigid Body

I tried to make up a rigid body simulation (a torus falling onto a plane) and I have tried to modify the bounciness from 0 up to 400, but I didn't observe any difference. What this parameter is supposed to do and how do I recognize the effect of the parameter ?

The setting controls how much energy the surface removes from a collision. The parameter can vary between 0.0 and 1.0 (you shouldn't set it outside that range) and a setting of 0.0 indicates no bounciness (all energy is removed). Increasing it up to 1.0 allows energy to remain after the collision.

Note that the collision energy is affected by both objects taking part in the collision so if either of the objects is set to 0.0 then no energy remains and there will be no 'bounce'. The Bounciness from both objects can be multiplied together to give a value of bounciness resulting in those objects colliding.

For example, for two objects each with a bounciness of 1.0, the result of a collision will have 1.0 x 1.0 = 1.0 (ie, all) of the energy before the collision. If either is zero then the result will be zero energy of their initial motion after the collision. For in-between values, the same holds - eg, if both have Bounciness of 0.8 then 0.8 x 0.8 = 0.64 of the energy will remain after the collision - ie, a ball would bounce 64% of the way back up from its starting height if it fell onto the surface.

In your case you likely have the other surface set to its default value of 0.0, resulting in no bounce (since anything multiplied by 0.0 is 0.0).

• Oh, I can see the point now: I have to set the other object bounciness also different from zero, right ? What did you mean by your **As an example, ** ? – user3357120 Nov 11 '20 at 17:34
• That's exactly it. Sorry about the 'As an example' text - left behind after editting. I've fixed it now. – Rich Sedman Nov 11 '20 at 17:36
• if both are 0.8 the result will be 0.64 only if hited directly straight onto the plane, i.e. not rotated ? – user3357120 Nov 11 '20 at 17:45
• Yes - if a ‘straight-on’ collision (the collider is travelling along the surface normal). If a more glancing blow then obviously the ‘bounce’ will be affected (I presume it would be scaled based on the cosine of the angle with the normal... if my understanding of the physics is correct). – Rich Sedman Nov 11 '20 at 17:48


text:
# Problem R*5.1 - Equivalence of twos forms of 2nd exactness condition

From Mtg 22-6

## Given

 \displaystyle \begin{align} g_{0}-\frac{dg_1}{dx}+ \frac{d^2g_2}{dx^2}=0 \end{align} (1.1)
 \displaystyle \begin{align} g_0=\frac{ \partial G}{\partial y} \end{align}
 \displaystyle \begin{align} g_1=\frac{\partial G}{\partial y'} \end{align}
 \displaystyle \begin{align} g_2=\frac{ \partial^2 G}{\partial y''} \end{align}

## Find

Equivalence of the form of 2nd exactness condition:

 \displaystyle \begin{align} f_{xx}+2pf_{xy}+p^2f_{yy}-g_{px}-pg_{py}+g_y+( f_{xp}+pf_{yp}+2f_y-g_{pp} )q=0 \end{align}

## Solution

- Solved on our own

Let's assume that p=y' and q=y.

 \displaystyle \begin{align} g_0&=\frac{ \partial G}{\partial y} \\ &=\frac{\partial( g+fy'' )}{\partial y} &=g_y+f_yq \end{align} (1.1)
 \displaystyle \begin{align} g_1&=\frac{ \partial G}{\partial y'} \\ &=\frac{\partial( g+fy'' )}{\partial p} &=g_p+f_pq \end{align}
 \displaystyle \begin{align} \frac{dg_1}{dx}&=\frac{d( g_p+f_pq )}{ dx } \\ &=\frac{\partial(g_p+f_pq)}{\partial x} +\frac{\partial(g_p+f_pq)}{\partial y}\frac{dy}{dx} +\frac{\partial(g_p+f_pq)}{\partial p}\frac{dp}{dx} \\ &=g_{px}+f_{px}q+(g_{py}+f_{yy}q)p+(g_{pp}+f_{yp}q)q \\ &=g_{px}+g_{py}p+g_{pp}q + f_{px}q +f_{yy}qp+f_{yp}q^2 \end{align} (1.2)
 \displaystyle \begin{align} g_2&=\frac{ \partial G}{\partial y''} \\ &=\frac{\partial( g+fy'' )}{\partial q} &=f \end{align}
 \displaystyle \begin{align} \frac{dg_2}{dx}&=\frac{d}{dx} (\frac{d( f )}{ dx }) \\ &=\frac{d}{dx} (f_x+f_yp+f_pq) \\ &=\frac{\partial(f_x+f_yp+f_pq)}{\partial x} +\frac{\partial(f_x+f_yp+f_pq)}{\partial y}\frac{dy}{dx} +\frac{\partial(f_x+f_yp+f_pq)}{\partial p}\frac{dp}{dx} \\ &=f_{xx}+f_{xy}p+f_{xp}q+f_{yx}p+f_{yy}p^2+f_{yp}pq+f_yq+f_{px}q+f_{py}pq+f_{pp}q^2\\ &=f_{xx}+2f_{xy}p+f_{yy}p^2+2f_{xp}q+2f_{yp}pq+f_{pp}q^2 \end{align} (1.3)

As a result,

 \displaystyle \begin{align} g_{0}-\frac{dg_1}{dx}+ \frac{d^2g_2}{dx^2}&=g_y+f_yq - (g_{px}+g_{py}p+g_{pp}q + f_{px}q +f_{yy}qp+f_{yp}q^2) + f_{xx}+2f_{xy}p+f_{yy}p^2+2f_{xp}q+2f_{yp}pq+f_{pp}q^2 \\ &=f_{xx}+2pf_{xy}+p^2f_{yy}-g_{px}-pg_{py}+g_y+( f_{xp}+pf_{yp}+2f_y-g_{pp} )q=0 \end{align}

# Problem R*5.2 - Verification of the exactness of Legendre and Hermite equations

From Mtg 27-1

## Given

Legendre equation:

 $\displaystyle G=(1-x^2) y^{''} - 2x y^{\prime} + n(n+1)y=0$ (2.1)

Hermite equation:

 $\displaystyle y^{''}-2xy^{\prime}+2ny = 0$ (2.2)

## Find

1. Verify the exactness of the designated L2-ODE-VC (Eq. 2.1, 2.2).
2. If Hermite equation is not exact, check whether it is in power form, and see whether it can be made exact using IFM with $\displaystyle h(x,y)=x^m y^n$
3. The first few Hermite polynomials are given as below.

 \displaystyle \begin{align} H_0(x) &=1 \\ H_1(x) &= 2x \\ H_2(x) &= 4x^2-2 \end{align} (2.3)

Verify that the equations in (2.3) are homogeneous solutions of the Hermite differential equation (Eq. 2.2).

## Solution

- Solved on our own

### Part 1

Legendre equation:
The first exactness condition is satisfied since the equation has a form of Eq. (2.4).

 \displaystyle \begin{align} G(x,y,y^{\prime}, y^{''}) &= g(x,y,p)+f(x,y,p)y^{''} \\ &= \underbrace{(1-x^2)}_{f(x,y,p)} y^{''} \underbrace{- 2x y^{\prime} + n(n+1)y}_{g(x,y,p)} \end{align} (2.4)

Method 1 : the 2nd exactness condition
In order to verify the second exactness condition, the following terms are computed.

 \displaystyle \begin{align} &f_{xx} = -2 \\ &f_{xy} = 0 \\ &f_{yy} = 0 \\ &g_{xp} = -2 \\ &g_{yp} = 0 \\ &g_y = n(n+1) \\ &f_{xp} = 0 \\ &f_y = 0 \\ &f_{yp} = 0 \\ &g_{pp} = 0 \end{align} (2.5)
 \displaystyle \begin{align} f_{xx}+2p f_{xy} + p^2 f_{yy} &= g_{xp} + p g_{yp} - g_y \\ -2 + 2p \cdot 0 + p^2 \cdot 0 &= -2 + p \cdot 0 - n(n+1) \\ \end{align} (2.6)
 \displaystyle \begin{align} f_{xp}+p f_{yp} + 2 f_y &= g_{pp} \\ 0 + p \cdot 0 + 2 \cdot 0 &= 0 \end{align} (2.7)
- The second exactness condition is satisfied when $\displaystyle n=0$ or $\displaystyle n=-1$.

Method 2 : the 2nd exactness condition
Following is another condition to prove the second exactness.

 $\displaystyle g_0-\frac{dg_1}{dx}+\frac{d^2 g_2}{dx^2}=0$ (2.8)

Using the definition of $\displaystyle g_i := \frac{\partial G}{\partial y ^{(i)}}$, the followings are computed.

 \displaystyle \begin{align} g_0&=\frac{\partial G}{\partial y^{(0)}} = n(n+1) \\ g_1&=\frac{\partial G}{\partial y^{(1)}} = -2x \\ g_2&=\frac{\partial G}{\partial y^{(2)}} = 1-x^2 \\ & \frac{dg_1}{dx}=-2 \\ & \frac{d^2 g_2}{dx^2} = -2 \end{align} (2.9)

After substituting the terms in Eq. (2.9), Eq. (2.8) becomes as following.

 \displaystyle \begin{align} n(n+1) - (-2) -2&= 0 \\ n(n+1) &= \end{align} (2.10)
- The second exactness condition is satisfied when $\displaystyle n=0$ or $\displaystyle n=-1$.

Hermite equation:
The first exactness condition is satisfied since the equation has a form of Eq. (2.8).

 \displaystyle \begin{align} G(x,y,y^{\prime}, y^{''}) &= g(x,y,p)+f(x,y,p)y^{''} \\ &= \underbrace{1}_{f(x,y,p)} \cdot y^{''} \underbrace{- 2x y^{\prime} + 2ny}_{g(x,y,p)} \end{align} (2.11)

Method 1 : the 2nd exactness condition
In order to verify the second exactness condition, the following terms are computed.

 \displaystyle \begin{align} &f_{xx} = 0 \\ &f_{xy} = 0 \\ &f_{yy} = 0 \\ &g_{xp} = -2 \\ &g_{yp} = 0 \\ &g_y = 2n \\ &f_{xp} = 0 \\ &f_y = 0 \\ &f_{yp} = 0 \\ &g_{pp} = 0 \end{align} (2.12)
 \displaystyle \begin{align} f_{xx}+2p f_{xy} + p^2 f_{yy} &= g_{xp} + p g_{yp} - g_y \\ 0 + 2p \cdot 0 + p^2 \cdot 0 &= -2 + p \cdot 0 - 2n \\ \end{align} (2.13)
 \displaystyle \begin{align} f_{xp}+p f_{yp} + 2 f_y &= g_{pp} \\ 0 + p \cdot 0 + 2 \cdot 0 &= 0 \end{align} (2.14)
- The second exactness condition is satisfied only when $\displaystyle n=-1$, but the condition is not satisfied if $\displaystyle n \neq -1$.

Method 2 : the 2nd exactness condition
Following is another condition to prove the second exactness.

 $\displaystyle g_0-\frac{dg_1}{dx}+\frac{d^2 g_2}{dx^2}=0$ (2.15)

Using the definition of $\displaystyle g_i := \frac{\partial G}{\partial y ^{(i)}}$, the followings are computed.

 \displaystyle \begin{align} g_0&=\frac{\partial G}{\partial y^{(0)}} = 2n \\ g_1&=\frac{\partial G}{\partial y^{(1)}} = -2x \\ g_2&=\frac{\partial G}{\partial y^{(2)}} = 1 \\ & \frac{dg_1}{dx}=-2 \\ & \frac{d^2 g_2}{dx^2} = 0 \end{align} (2.16)

After substituting the terms in Eq. (2.16), Eq. (2.15) becomes as following.

 \displaystyle \begin{align} 2n - (-2) +0&= 0 \\ 2(n+1) &= \end{align} (2.17)
- The second exactness condition is satisfied only when $\displaystyle n=-1$.

### Part 2

As shown in the Eq. (2.10), the Hermite equation does not satisfy the second exactness condition if n is a positive value. To see if the equation can be made exact using IFM with $\displaystyle h(x,y) = x^m y^n$, multiply the Eq. (2.2) by $\displaystyle h(x,y)$. Also, the term n in the Eq. (2.2) is replaced with $\displaystyle \lambda$ to avoid confusion.

 $\displaystyle x^m y^n ( y^{''} - 2x y^{\prime} + 2\lambda y ) = 0$ (2.18)
 $\displaystyle \underbrace{x^m y^n}_{f(x,y,p)} y ^{''} \underbrace{-2x^{m+1}y^n y^{\prime}+2\lambda x^{m}y^{n+1}}_{g(x,y,p)} = 0$ (2.19)

It is assumed that the Eq. (2.13) is exact. Then, the integrating factor $\displaystyle h(x,y)$ can be obtained by utilizing the second exactness condition as following.

 \displaystyle \begin{align} &f_{xx} = m(m-1)x^{m-2}y^n \\ &f_{xy} = mnx^{m-1}y^{n-1} \\ &f_{yy} = n(n-1)x^my^{n-2} \\ &g_{xp} = -2(m+1)x^my^n \\ &g_{yp} = -2nx^{m+1} y^{n-1} \\ &g_y = -2nx^{m+1} y^{n-1}p + 2(n+1)\lambda x^m y^n \\ &f_{xp} = 0 \\ &f_y = nx^m y^{n-1} \\ &f_{yp} = 0 \\ &g_{pp} = 0 \end{align} (2.20)

With the terms computed above, the two equations below which represent the second exactness condition are exploited.

 $\displaystyle f_{xx}+2p f_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_y$ (2.21)
 $\displaystyle f_{xp}+p f_{yp} + 2 f_y = g_{pp}$ (2.22)

First, substitute the terms in the Eq. (2.14) into the Eq. (2.16).

 \displaystyle \begin{align} f_{xp}+p f_{yp} + 2 f_y &= g_{pp} \\ 0 + p \cdot 0 + 2nx^m y^{n-1} &= 0 \end{align} (2.23)

The only way the Eq. (2.17) is satisfied is to set n equal to zero.

 $\displaystyle \therefore n=0$ (2.24)

With knowing n=0, substitute the terms in the Eq. (2.14) into the Eq. (2.15).

 \displaystyle \begin{align} f_{xx}+2p f_{xy} + p^2 f_{yy} &= g_{xp} + p g_{yp} - g_y \\ m(m-1)x^{m-2} &= -2(m+1)x^m - 2 \lambda x^m \\ &= -2 \left(m+1+\lambda \right) x^m \end{align} (2.25)

The both sides of the Eq. (2.19) are equal only when the coefficients of $\displaystyle x^{m-1}, x^m$ are zero. Then the following is concluded.

 \displaystyle \begin{align} m(m-1) &= 0 \\ m+1+ \lambda &= 0 \end{align} (2.26)

The first equation in the Eq. (2.20) implies that $\displaystyle m=0$ or $\displaystyle m=1$. Then $\displaystyle \lambda$ is determined using the second one in the Eq. (2.20).

 \displaystyle \begin{align} &\lambda = -1 \mbox{ when } m = 0 \\ &\lambda = -2 \mbox{ when } m = 1 \end{align} (2.27)

Since n is determined to be zero already, m can not be zero. Hence, m is chosen to be one and $\displaystyle \lambda$ is 2.

 \displaystyle \begin{align} n &= 0 \\ m &= 1 \\ \lambda &= -2 \end{align} (2.28)
 \displaystyle \begin{align} \therefore h(x,y) &= x^1 y^0 \\ &= x \end{align} (2.29)

Here, the exactness of the Hermite equation with $\displaystyle h(x,y)$ is justified.

 \displaystyle \begin{align} G(x,y,y^{\prime}, y^{''}) &= g(x,y,p)+f(x,y,p)y^{''} \\ &= xy^{''} - 2x^2 y^{\prime} -4xy \end{align} (2.30)

In order to prove the second exactness of the Hermite equation, the Method 2 is used.

 $\displaystyle g_0-\frac{dg_1}{dx}+\frac{d^2 g_2}{dx^2}=0$ (2.31)

Using the definition of $\displaystyle g_i := \frac{\partial G}{\partial y ^{(i)}}$, the followings are computed.

 \displaystyle \begin{align} g_0&=\frac{\partial G}{\partial y^{(0)}} = -4x \\ g_1&=\frac{\partial G}{\partial y^{(1)}} = -2x^2 \\ g_2&=\frac{\partial G}{\partial y^{(2)}} = x \\ & \frac{dg_1}{dx}=-4x \\ & \frac{d^2 g_2}{dx^2} = 0 \end{align} (2.32)

After substituting the terms in Eq. (2.32), Eq. (2.31) becomes as following.

 $\displaystyle -4x-(-4x)+0=0$ (2.33)
- Hermite equation with IFM $\displaystyle h(x,y) = x$ is exact.

### Part 3

Hermite differential equation is known to have a polynomial solution, which has following form.

 $\displaystyle y(x) = \sum_{n=0}^{\infty} a_n x^n$ (2.34)

Compute the first and the second derivative of Eq. (2.34) with respect to x,

 \displaystyle \begin{align} y^{\prime}(x) &= \sum_{n=1}^{\infty}n a_n x^{n-1} \\ y^{''}(x) &= \sum_{n=2}^{\infty}n(n-1) a_n x^{n-2} \\ \end{align} (2.35)

Substitute Eq. (2.34), (2.35) into Eq. (2.2),

 \displaystyle \begin{align} &\sum_{n=2}^{\infty}n(n-1)a_n x^{n-1} - 2x \sum_{n=1}^{\infty} n a_n x^{n-1} + 2 \lambda \sum_{n=1}^{\infty} a_n x^n = 0 \\ &\sum_{n=2}^{\infty}n(n-1)a_n x^{n-1} - \sum_{n=1}^{\infty} 2n a_n x^{n} + \sum_{n=1}^{\infty} 2 \lambda a_n x^n = 0 \\ \end{align} (2.36)

Shift the first summation up by two units,

 $\displaystyle \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2} x^{n} - \sum_{n=1}^{\infty} 2n a_n x^{n} + \sum_{n=0}^{\infty} 2 \lambda a_n x^n = 0$ (2.37)

The second summation starts at n=1, while the other summations start at n=0. Following justifies the second summation starting at n=0.

 \displaystyle \begin{align} \sum_{n=1}^{\infty} 2n a_n x^n &= 2 \cdot 0 \cdot a_0 x^0 + \sum_{n=1}^{\infty}2n a_n x^n \\ &= \sum_{n=0}^{\infty}2n a_n x^n \end{align} (2.38)

Combine all summations in Eq. (2.37),

 $\displaystyle \sum_{n=0}^{\infty} \left[ (n+2)(n+1)a_{n+2} - 2n a_n + 2 \lambda a_n \right] x^n = 0$ (2.39)

Hence, the following recurrence relation is obtained from Eq. (2. 39).

 $\displaystyle (n+2)(n+1)a_{n+2} - 2n a_n + 2 \lambda a_n = 0 \mbox{ for all } n=0,1,2,3, \cdots$ (2.40)

Simplify Eq. (2.40),

 $\displaystyle a_{n+2} = \frac{2(n-\lambda)}{(n+2)(n+1)} a_n \mbox{ for all } n=0,1,2,3, \cdots$ (2.41)

Using the recurrence relation in Eq. (2.41), the Hermite polynomials given in the problem are obtained.
Let the initial conditions and $\displaystyle \lambda$ be as follows.

 $\displaystyle y(0)=a_0=1, \mbox{ } y^{\prime}(0)=a_1=0, \mbox{ } \lambda = 0$ (2.42)

Then,

 \displaystyle \begin{align} a_2 &= \frac{2(0-0)}{2 \cdot 1} a_0 = 0 \\ a_3 &= \frac{2(1-0)}{3 \cdot 2} a_1 = 0 \\ a_4 &= a_5 = a_6 = a_7 = \ldots = 0 \end{align} (2.43)
- Hence, Hermite polynomial is $\displaystyle H_0(x) = 1$ in this case.

Let the initial conditions and $\displaystyle \lambda$ be as follows.

 $\displaystyle y(0)=a_0=0, \mbox{ } y^{\prime}(0)=a_1=2, \mbox{ } \lambda = 1$ (2.44)

Then,

 \displaystyle \begin{align} a_2 &= \frac{2(0-1)}{2 \cdot 1} a_0 = 0 \\ a_3 &= \frac{2(1-1)}{3 \cdot 2} a_1 = 0 \\ a_4 &= a_5 = a_6 = a_7 = \ldots = 0 \end{align} (2.45)
- Hence, Hermite polynomial is $\displaystyle H_1(x) = 2x$ in this case.

Let the initial conditions and $\displaystyle \lambda$ be as follows.

 $\displaystyle y(0)=a_0=-2, \mbox{ } y^{\prime}(0)=a_1=0, \mbox{ } \lambda = 2$ (2.46)

Then,

 \displaystyle \begin{align} a_2 &= \frac{2(0-2)}{2 \cdot 1} a_0 = 4 \\ a_3 &= \frac{2(1-2)}{3 \cdot 2} a_1 = 0 \\ a_4 &= \frac{2(2-2)}{4 \cdot 3} a_2 = 0 \\ a_5 &= a_6 = a_7 = a_8 = \ldots = 0 \end{align} (2.47)
- Hence, Hermite polynomial is $\displaystyle H_2(x) = 4x^2-2$ in this case.

# References

• Hermite Polynomial, WolframMathworld [1]

# Problem R*5.3 Method of undetermined coefficients

From Mtg 30-3

## Given

Given L4-ODE-CC,

$\displaystyle X^{(4)} - K^4 X = 0$

(3.1)

Assuming,

$\displaystyle X(x) = e^{(rx)}$

(3.2)

and substituting this in Eq. (3.1), we get 4 solutions for r,

$\displaystyle r_{1,2} = \pm K$

$\displaystyle r_{3,4} = \pm i \, K$

(3.3)

## Find

Expressions for X(x) in terms of $\displaystyle \cos K x, \ \sin K x, \ \cosh K x, \ \sinh K x$

## Solution

- Solved on my own

We can write final solution X(x) as,

$\displaystyle X(x) = \sum_{i=1}^4 C_i e^{(r_i x)}$

(3.4)

Substituting all the ri values from Eq. (3.3),

$\displaystyle X(x) = C_1 e^{K x} + C_2 e^{-K x} + C_3 e^{iK x} + C_4 e^{-iK x}$

(3.5)

Expanding the exponential terms into sin, cos, sinh and cosh terms, we get,

$\displaystyle X(x) = C_1 \cosh K x \ + C_1 \sinh K x \ + C_2 \cosh K x \ - C_2 \sinh K x \ + C_3 \cos K x \ + i C_3 \sin K x \ + C_4 \cos k x - \ i C_4 \sin k x$

(3.6)

for X(x) to be real, imaginary part in above Eq. (3.6) must be real, i.e. (C3 - C4) must be imaginary

$\displaystyle C_3 - C_4 = ib_4$

and, (C3 + C4) must be real,

$\displaystyle C_3 + C_4 = b_3$

Therefore, C3 and C4 will be given by,

$\displaystyle C_3 = \frac{1}{2}(b_3 + ib_4)$

$\displaystyle C_4 = \frac{1}{2}(b_3 - ib_4)$

This means C3 and C4 given above are complex conjugates.

Therefore, we get final solution to be,

$\displaystyle X(x) = (C_1 + C_2) \cosh K x + (C_1 - C_2) \sinh K x + b_3 \cos K x - b_4 \sin K x$

(3.7)

# Problem R*5.4 - Find $\displaystyle y_{xxxxx}$ in terms of the derivatives of y with respect to t

From Mtg 31-1

## Given

In Euler L2-ODE-VC, suppose: $\displaystyle x=e^t$.

## Find

$\displaystyle y_{xxxxx}$

## Solution

- Solved on my own

Since:

 $\displaystyle \frac{dt}{dx} = \left( \frac{dx}{dt} \right)^{-1} = \left( \frac{d(e^t)}{dt} \right)^{-1} = e^{-t}$ (4.1)

So:

 $\displaystyle y_x = \frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = e^{-t} y_t$ (4.2)
 \displaystyle \begin{align} y_{xx} &= \frac{dy_x}{dx} \\ &= \frac{dy_x}{dt} \frac{dt}{dx}\\ &= e^{-2t} (y_{tt}-y_t)\\ \end{align} (4.3)
 \displaystyle \begin{align} y_{xxx} &= \frac{dy_{xx}}{dx} \\ &= \frac{dy_{xx}}{dt} \frac{dt}{dx}\\ &= e^{-3t} (y_{ttt}-3y_{tt}+2y_t)\\ \end{align} (4.4)
 \displaystyle \begin{align} y_{xxxx} &= \frac{dy_{xxx}}{dx} \\ &= \frac{dy_{xxx}}{dt} \frac{dt}{dx}\\ &= e^{-4t} (y_{tttt}-6y{ttt}+11y_{tt}-6y_t)\\ \end{align} (4.5)
 \displaystyle \begin{align} y_{xxxxx} &= \frac{dy_{xxxx}}{dx} \\ &= \frac{dy_{xxxx}}{dt} \frac{dt}{dx}\\ &= e^{-t} \frac{d \left( e^{-4t}(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t) \right)}{dt}\\ &= e^{-5t} (y_{ttttt}-10y_{tttt}+35y_{ttt}-50y_{tt}+24y_t) \end{align} (4.6)

# Problem R*5.5 - Solve y and plot the solution

From Mtg 31-3

## Given

Euler L2-ODE-VC:

 $\displaystyle x^2 y''-2xy'+2y=0$ (5.1)

and trial solution:

 $\displaystyle y=x^r$ (5.2)

with boundary conditions:

 $\displaystyle \begin{cases} y(1)=-4\\ y(2)=7\\ \end{cases}$ (5.3)

## Find

The solution $\displaystyle y(x)$ and plot it.

## Solution

- Solved on my own

Substitute $\displaystyle x^r$ for $\displaystyle y$ in equation (5.1):

 $\displaystyle \underbrace{x^r}_{\displaystyle \ne 0} (r^2-3r+2)=0$ (5.4)

So the characteristic equation:

 $\displaystyle r^2-3r+2=0 \Rightarrow \begin{cases} r_1=1\\ r_2=2\\ \end{cases}$ (5.5)

So the solution is:

 $\displaystyle y(x)=c_1 x + c_2 x^2$ (5.6)

Use the boundary conditions:

 $\displaystyle \begin{cases} y(1)=c_1+c_2=-4\\ y(2)=2c_1+4c_2=7\\ \end{cases} \Rightarrow \begin{cases} c_1=- \frac{23}{2}\\ c_2= \frac{15}{2}\\ \end{cases}$ (5.7)

So the solution is:

 $\displaystyle y(x)= \frac{15}{2} x^2 - \frac{23}{2} x$ (5.8)

Plot $\displaystyle y(x)$.

Matlab Code:

 clear all close all clc %homogeneous solutions y=@(x) (15/2)*(x^2)-(23/2)*x; %plotting hold on fplot(y,[-7 7],'r'); title('Plotting'); xlabel('x'); ylabel('y(x)'); legend('y=(15/2)*(x^2)-(23/2)*x')

# Problem R5.6: Equivalence of Two methods to solve Euler Ln-ODE-VC

From Mtg 31-5

## Given

Given Euler Ln-ODE-VC,

$\displaystyle \sum_{i=0}^n a_i x^i y^{(i)} = 0$

(6.1)

where, $\displaystyle y^{(i)}$ is ith derivative of y w.r.t. x,

Two methods of solving above equation,

Method 1:

Stage 1: Transformation of variables

$\displaystyle x = e^t$

(6.2)

Stage 2: Trial Solution

$\displaystyle y = e^{rt} ,\ {\color{red} r = \text{constant}}$

(6.3)

Method 2: Trial Solution

$\displaystyle y = x^{r} ,\ {\color{red} r = \text{constant}}$

(6.4)

## Find

Show equivalence of methods 1 and 2 shown above in Eq. (6.2), (6.3) and Eq. (6.4) respectively

## Solution

- Solved on our own

Let's take method - 2 shown in Eq. (6.4), first. Differentiating Eq. (6.4) w.r.t. x, we get

$\displaystyle y_x = y' = r x^{r-1}$

(6.5)

For, ith derivative of y w.r.t. x, i.e. y(i), we have,

$\displaystyle y^{(i)} = r(r-1)...(r-i+1)x^{(r-1)}$

(6.6)

Substituting Eq. (6.6), i.e. definition of y(i) in Eq. (6.1), we get,

$\displaystyle \sum_{i=0}^n a_i x^i r(r-1)...(r-i+1) x^{(r-i)} = 0$

(6.7a)

$\displaystyle \sum_{i=0}^n a_i x^r r(r-1)...(r-i+1) = 0$

(6.7b)

Eliminating xr, which is a common factor in the all the terms of above summation, we get,

$\displaystyle \sum_{i=0}^n a_i r(r-1)...(r-i+1) = 0$

(6.7c)

So, we solve the above equation for r, and that provides us solution for Eq. (6.1).

Now, for method - 1, we will try to get the same form as in Eq. (6.7c)

Stage 1: Substitute x = et in Eq. (6.1), we write yx as,

$\displaystyle y_x = y_t \frac {dt}{dx} = e^{-t} y_t$

(6.8)

Stage 2: Substitute y = ert,

Using above we can write yt as, r*ert. Substitute this into Eq. (6.8), and we get,

$\displaystyle y^{(1)} = y_x = re^{rt}e^-t = r e^{(r-1)t}$

(6.9)

Performing another differentiation of above equation to get yxx, we get,

$\displaystyle y^{(2)} = y_{xx} = \frac {d}{dx} (y_x) = \frac {d}{dx} (r e^{(r-1)t}) = \frac {d}{dt} (r e^{(r-1)t}) \frac {dt}{dx} = r(r-1) e^{(r-1)t} e^{-t} = r(r-1)e^{(r-2)t}$

(6.10)

We can see that as done above for y(2), we can write in general for y(i), we can write,

$\displaystyle y^{(i)} = r(r-1)...(r-i+1) e^{(r-i)t}$

(6.10)

Substituting definitions of both x and y as seen in Eq. (6.2) and Eq. (6.10) respectively, into Eq. (6.1) we get,

$\displaystyle \sum_{i=0}^n a_i e^{it} r(r-1)...(r-i+1) e^{(r-i)t} = 0$

(6.11a)

Above can be simplified to following form very easily (combining both the exponential terms in above expression),

$\displaystyle \sum_{i=0}^n a_i e^{rt} r(r-1)...(r-i+1) = 0$

(6.11b)

Since, e(rt) is not dependent on i, we can take this term out of the summation as a common factor, and write rest of the terms as,

$\displaystyle \sum_{i=0}^n a_i r(r-1)...(r-i+1) = 0$

(6.11c)

Above Eq. (6.11c) is exactly same as Eq. (6.7c), and therefore it proves that by both methods 1 & 2 we will get same values for r,
and hence both methods are equivalent.

# Problem R*5.7-Euler L2-ODEs

From Mtg 32-1

## Given

Characteristic equation

 \displaystyle \begin{align} (r - \lambda)^2=0 \end{align} (7.1)
 \displaystyle \begin{align} a_2x^2y''+a_1xy'+a_0y=0 \end{align} (7.2)
 \displaystyle \begin{align} b_2y''+b_1y'+b_0y=0 \end{align} (7.3)

## Find

1.1) Find $a_2, a_1, a_0$ such that (7.1) is characteristic equation of (7.2)

1.2) 1st homogeneous solution : $y_1(x)=x^\lambda$

1.3) Complete solution : Find c(x) such that $y(x)=c(x)y_1(x)$

1.4) Find the 2nd homogeneous solution $y_2(x)$

2.1) Find $b_2, b_1, b_0$ such that (7.1) is characteristic equation of (7.3)

2.2) 1st homogeneous solution

2.3) Complete solution : Find c(x) such that $y(x)=c(x)y_1(x)$

2.4) Find the 2nd homogeneous solution $y_2(x)$

## Solution

- Solved on our own

### Part 1.1

Assume that the trial solution is following.

 \displaystyle \begin{align} y=x^r \end{align} (7.4)

Then, compute the first and the second derivatives of the trial solution,

 \displaystyle \begin{align} y'=rx^{r-1} \end{align} (7.5)
 \displaystyle \begin{align} y''=r(r-1)x^{r-2} \end{align} (7.6)

Substitute the equation (7.4), (7.5), (7.6) into (7.2) and simplify the equation,

 \displaystyle \begin{align} a_2x^2r(r-1)x^{r-2}+a_1xrx^{r-1}+a_0x^r=0 \\ a_2r(r-1)x^r+a_1rx^r+a_0x^r=0 \\ \left[ a_2r(r-1)+a_1r+a_0 \right] x^r=0 \end{align} (7.7)

Then, we get

 \displaystyle \begin{align} a_2r^2+(a_1-a_2)r+a_0=0 \end{align} (7.8)

Compare (7.8) to (7.1),

 \displaystyle \begin{align} a_2r^2+(a_1-a_2)r+a_0 &= (r-\lambda)^2 \\ &= r^2-2r\lambda+\lambda^2 \\ \end{align} (7.9)

Comparing the coefficients of LHS and RHS,

 \displaystyle \begin{align} a_2&=1 \\ a_1&=1-2\lambda \\ a_0&=\lambda^2 \end{align} (7.10)

### Part 1.2

From the characteristic equation in Eq. (7.1), the solution of the characteristic equation is determined as follows.

 $\displaystyle r = \lambda$ (7.11)

Hence, the 1st homogeneous solution can be

 \displaystyle \begin{align} y_1=x^\lambda \end{align} (7.12)

### Part 1.3

 \displaystyle \begin{align} y=c(x)y_1(x) \end{align} (7.13)

Compute the first and the second derivatives of Eq. (7.13),

 \displaystyle \begin{align} y^{\prime}(x)&=c^{\prime}(x)y_1(x)+c(x)y_1^{\prime}(x) \\ y^{''}(x) &=c^{''}(x)y_1(x)+c^{\prime}(x)y_1^{\prime}(x)+c^{\prime}(x)y_1^{\prime}(x)+C(x)y_1^{''}(x) \end{align} (7.14)

Substitute (7.13) and (7.14) into (7.2) and simplify the equation,

 \displaystyle \begin{align} a_2x^2\left[c^{''}(x)y_1(x)+c^{\prime}(x)y_1^{\prime}(x)+c^{\prime}(x)y_1^{\prime}(x)+c(x)y_1^{''}(x)\right] + a_1x\left[ c^{\prime}(x)y_1(x)+c(x)y_1^{\prime}(x) \right] +a_0c(x)y_1(x)&=0 \\ c(x) \cancelto{0}{\left[ a_2x^2y_1^{''}(x) + a_1xy_1^{\prime}(x)+a_0y_1(x) \right]} + c^{\prime}(x) \left[ 2a_2x^2y_1^{\prime}(x) + a_1xy_1(x) \right] + c^{''}(x)\left[ a_2x^2y_1(x) \right] &= 0 \\ \end{align} (7.15)

The first term in Eq. (7.15) cancels out using Eq. (7.2). Then,

 \displaystyle \begin{align} c^{\prime}(x) \left[ 2a_2x^2y_1^{\prime}(x) + a_1xy_1(x) \right] + c^{''}(x)\left[ a_2x^2y_1(x) \right] = 0 \\ \end{align} (7.16)

Substitute coefficients and homogeneous solution,

 \displaystyle \begin{align} c^{\prime}(x) \left[ 2a_2x^2y_1^{\prime}(x) + a_1xy_1(x) \right] + c^{''}(x)\left[ a_2x^2y_1(x) \right] \\ \end{align}
 \displaystyle \begin{align} &= c^{\prime}(x) \left[ 2x^2 \lambda x^{ \lambda-1 } + 1-2\lambda x x^\lambda \right] + c^{''}(x) \left[ x^2 x^\lambda \right] \\ &= c^{\prime}(x) \left[ 2 \lambda x^{ \lambda+1 } + 1-2\lambda x^{\lambda+1} \right] + c^{''}(x) \left[ x^{\lambda+2} \right] \\ &= c^{\prime}(x) x^{\lambda+1} + c^{''}(x) x^{\lambda+2} \\ &= \left[ c^{\prime}(x) + c^{''}(x) x \right] x^{\lambda+1} =0 \\ \end{align}
 \displaystyle \begin{align} \therefore \left[ c^{\prime}(x) + c^{''}(x)x \right] = 0 \end{align} (7.17)

Let $\displaystyle c^{\prime}(x)=z(x)$,

 \displaystyle \begin{align} z(x)+z'(x)x &=0 \\ z+x\frac{dz}{dx}&=0 \\ z&=-x\frac{dz}{dx} \\ \frac{1}{z}dz &= -\frac{1}{x}dx \\ \int \frac{1}{z}dz &= \int \frac{1}{x}dx \\ \end{align} (7.18)
 $\displaystyle \log (z)= \log (x) + C_1$ (7.19)
 \displaystyle \begin{align} z&= e^{-(\log (x)+C_1)} \\ &= e^ {\log (x {e^{C_1}})^{-1}} \\ &= \frac{1}{e^{C_1}x} \\ &=C_2\frac{1}{x} \mbox{ } (C_2=e^{-C_1}) \end{align} (7.20)

Then, compute the function $\displaystyle c(x)$,

 \displaystyle \begin{align} c(x)&=\int z dx \\ &=\int C_2 \frac{1}{x} dx \\ &=C_2(\log(x)+C_3) \end{align} (7.21)

As a result,

 \displaystyle \begin{align} y(x)&=C_2(\log(x)+C_3)y_1(x) \\ &=C_2x^\lambda \log(x)+ C_4x^\lambda \\ &=C_1^* x^\lambda \log(x) + C_2^*x^\lambda \end{align} (7.22)

### Part 1.4

From 1.3, the general homogeneous solution is

 \displaystyle \begin{align} y(x)&=C_1^* x^\lambda \log(x) + C_2^*x^\lambda \end{align}

From (7.12), $\displaystyle y_1(x)$ is $\displaystyle x^\lambda$.

So, $\displaystyle y_2(x)$ is

 \displaystyle \begin{align} y_2(x)&= x^\lambda \log(x) \end{align}

### Part 2.1

The trial solution for the Euler equation with constant coefficient:

 \displaystyle \begin{align} y=e^{xr} \end{align} (7.23)

Compute the first and the second derivatives

 \displaystyle \begin{align} y^{\prime}(x)=re^{xr} \end{align} (7.24)
 \displaystyle \begin{align} y^{''}(x)=r^2e^{rx} \end{align} (7.25)

Substitute (7.23),(7.24),(7.25) into (7.3) and simplify it,

 \displaystyle \begin{align} b_2r^2e^{rx}+b_1re^{rx}+b_0e^{rx} &=0 \\ \left[ b_2r^2+b_1r+b_0 \right] e^{rx} &=0 \\ \end{align}

Then,

 \displaystyle \begin{align} b_2r^2+b_1r+b_0=0 \\ \end{align}
 \displaystyle \begin{align} b_2r^2+b_1r+b_0&=(r-\lambda)^2 \\ &=r^2-2r\lambda+\lambda^2 \end{align}

Compare LHS and RHS,

 \displaystyle \begin{align} b_2=1 \end{align} (7.26)
 \displaystyle \begin{align} b_1=-2\lambda \end{align} (7.27)
 \displaystyle \begin{align} b_0=\lambda^2 \end{align} (7.28)

### Part 2.2

From (7.1), $\displaystyle r=\lambda$, As a result,

 \displaystyle \begin{align} y_1=e^{x\lambda} \end{align} (7.29)

### Part 2.3

 \displaystyle \begin{align} y=c(x)y_1(x) \end{align} (7.30)

Compute the first and the second derivatives

 \displaystyle \begin{align} y^{'}(x)=c^{'}(x)y_1(x)+c(x)y_1^{'}(x) \end{align} (7.31)
 \displaystyle \begin{align} y^{''}(x)=c^{''}(x)y_1(x)+c^{'}(x)y_1^{'}(x)+c^{'}(x)y_1^{'}(x)+c(x)y_1^{''}(x) \end{align} (7.32)

Substituting (7.25),(7.26) and (7.27) into (7.3)

 \displaystyle \begin{align} b_2\left[c^{''}(x)y_1(x)+c^{'}(x)y_1^{'}(x)+c^{'}(x)y_1^{'}(x)+C(x)y_1^{''}(x)\right] + b_1\left[ c^{'}(x)y_1(x)+C(x)y_1^{'}(x) \right] +b_0c(x)y_1(x) \\ c(x) \cancelto{0} {\left [ b_2y_1^{''}(x) + b_1y_1^{'}(x)+b_0y_1(x) \right]} + c^{'}(x) \left[ 2b_2y_1^{'}(x) + b_1y_1(x) \right] + c^{''}(x)\left[ b_2y_1(x) \right] = 0 \\ \end{align}

From (7.3),

 \displaystyle \begin{align} b_2y_1^{''}(x) + b_1y_1^{'}(x)+b_0y_1(x)=0 \end{align}

Then,

 \displaystyle \begin{align} c^{'}(x) \left[ 2b_2y_1^{'}(x) + b_1y_1(x) \right] + c^{''}(x)\left[ b_2y_1(x) \right] = 0 \\ \end{align}

Substituting coefficients and homogeneous solution,

 \displaystyle \begin{align} &c^{'}(x) \left[ 2b_2y_1^{'}(x) + b_1y_1(x) \right] + c^{''}(x)\left[ b_2y_1(x) \right] \\ &= c^{'}(x) \cancelto{0} {\left[ 2\lambda e^{ \lambda x } -2\lambda e^{\lambda x} \right]} + c^{''}(x) \left[ e^{\lambda x} \right] \\ &= c^{''}(x) e^{\lambda x} = 0 \\ \end{align}

Then,

 \displaystyle \begin{align} \therefore c^{''}(x)=0 \end{align} (7.33)

Solve (7.33),

 \displaystyle \begin{align} c^{'}(x) &= \int c^{''}(x) dx \\ &= \int 0 dx \\ &= C_1 \\ c^{'}(x)&=C_1 \\ c(x)&=\int c^{'}(x) dx\\ &=\int C_1 dx\\ &=C_1x+C_2 \end{align} (7.34)

As a result,

 \displaystyle \begin{align} y(x)&= c(x)y_1(x) \\ &= (C_1x+C_2)e^{\lambda x} \\ &= C_1xe^{\lambda x}+ C_2e^{\lambda x} \end{align} (7.35)

### Part 2.4

From 2.3, the general homogeneous solution is

 \displaystyle \begin{align} y(x)&=(C_1x+C_2)e^{\lambda x} \\ &= C_1xe^{\lambda x} + C_2e^{\lambda x} \\ \end{align}

From (7.29), $\displaystyle y_1(x)$ is $\displaystyle e^{\lambda x}$.

So, $\displaystyle y_2(x)$ is

 \displaystyle \begin{align} y_2(x)&= xe^{\lambda x} \end{align}

# Problem R*5.8 - Variation of parameters

From Mtg 32-2

## Given

A General non-homogenous L1-ODE-VC (Linear, 1st order, Ordinary Differential Equation with Variable Coefficient),

 $\displaystyle P(x)\,y' + Q(x)\,y = \underbrace{R(x)}_{\color{blue}{\neq 0}}$ (8.1)

which can be written in the form given below,

 $\displaystyle \underbrace{\color{blue}{1}}_{\color{blue}{a_1(x)}} \cdot \underbrace{y'}_{\color{blue}{y^{(1)}}} + \underbrace{\frac{Q(x)}{P(x)}}_{\color{blue}{a_0(x)}}y^{\color{blue}{(0)}} = \underbrace{\frac{R(x)}{P(x)}}_{\color{blue}{b(x)}}$ (8.2)

The form can be written as,

 $\displaystyle y' + a_0(x)y^{(0)} = b(x)$ (8.3)

## Find

Find the particular solution, $\displaystyle y_P(x)$ by the use of the method of variation of parameters after knowing the homogeneous solution $\displaystyle y_H(x)$, i.e., let $\displaystyle y(x) = A(x)y_H(x)$, with $\displaystyle A(x)$ being the unknown to be found.

## Solution

- Solved on our own

For particular solution, $\displaystyle y_P (x)$, the use variation of parameters requires $\displaystyle y_P(x)$ to be following,

 $\displaystyle y_p(x) = \sum_{i=1}^{n} c_i(x) y_i(x)$ (8.4)

for $n^{th}$ order non-linear homogenous ODE. For our case, n = 1, therefore we get,

 $\displaystyle y_p(x) = A(x) y_H(x)$ (8.5)

It is important to note for method of variation of parameters that above, $\displaystyle c_1$ is a function of $\displaystyle x$, and is determined by substituting the particular solution. Thus,

 $\displaystyle y_H^{'}(x) + y_P^{'}(x) + a_0(x)[y_H(x) + y_P(x)] = b(x)$ (8.6)

Substitute Eq. 8.5 into Eq. 8.6 to get,

 $\displaystyle y_H^{'}(x) + [A^{'}(x)y_H(x) + A(x)y_H^{'}(x)] + a_0(x)y_H(x) + a_0(x)[y_H(x) + A(x)y_H(x)] = b(x)$ (8.7)

and, rearrange the terms,

 $\displaystyle \underbrace{[y_H^{'}(x) + a_0(x)y_H(x)]}_{=0} + A^{'}(x)y_H(x) + A(x)\underbrace{[y_H^{'}(x) + a_0(x)y_H(x)]}_{=0} = b(x)$ (8.8)

Therefore $\displaystyle A(x)$ is given by,

 $\displaystyle A^{'}(x) = \frac{b(x)}{y_H(x)}$ (8.9)

Integrate both sides w.r.t. x,

 $\displaystyle A(x) = \int^x \frac{b(s)}{y_H(s)} ds$ (8.10)

Substituting for the expression of $\displaystyle y_H(x)$ from the homework problem 2.17 into above, we will get,

 $\displaystyle A(x) = \int^x \underbrace{\exp [ \int^s a_0(t)dt ]}_{=h(s)} b(s) ds$ (8.11)

Therefore particular solution can finally be written as,

 $\displaystyle y_p(x) = [\int^x h(s) b(s) ds] \frac{1}{h(s)}$ (8.12)

# Problem R*5.9 Special IFM Solution

From Mtg 32-7

## Given

 \displaystyle \begin{align} a_2x^2y''+a_1xy'+a_0y=0 \end{align} (9.1)

## Find

Special IFM to solve for general $\displaystyle f(t)$

## Solution

- Solved on our own

First, find integrating factor $\displaystyle h(t,y)$

 $\displaystyle h(t,y)[{a_2}y''+{a_1}y'+{a_0}y]=h(t,y)f(t)$ (Eq 9.2)

Suppose $\displaystyle h(t,y)$ is a function of $\displaystyle t$, thus $\displaystyle h(t,y)=h(t)$

Rearrange Eq 9.2, we have

 $\displaystyle h(t)[{a_2}y''+{a_1}y'+{a_0}y-f(t)]=0$ (Eq 9.3)
 $\displaystyle \frac{d\phi }{dt}=0$ (Eq 9.4)

Thus

 $\displaystyle F(t,y,p,p')=h(t){a_2}p'+h(t){a_1}p+h(t){a_0}y-h(t)f(t)=f(t,y,p)p'+g(t,y,p)$ (Eq 9.5)

Where $\displaystyle p:=y'$

According to the 2nd Exactness Condition

 $\displaystyle {f}_{tt}+2p{f}_{ty}+{p}^{2}{f}_{yy}={g}_{tp}+p{g}_{yp}-{g}_{y}$ (Eq 9.6)
 $\displaystyle {f}_{tp}+p{f}_{yp}+2{f}_{y}={g}_{pp}$ (Eq 9.7)

We have

 $\displaystyle f_{tt}={h_{tt}}{a_2}$ (Eq 9.8)
 $\displaystyle f_{ty}=0$ (Eq 9.9)
 $\displaystyle f_{yy}=0$ (Eq 9.10)
 $\displaystyle g_{tp}={h_{t}}{a_1}$ (Eq 9.11)
 $\displaystyle g_{yp}=0$ (Eq 9.12)
 $\displaystyle g_{y}={h}{a_0}$ (Eq 9.13)

Substituting Eq 9.8~9.13 in Eq 9.6, we get

 $\displaystyle h_{tt}a_2={h_{t}}{a_1}-h{a_0}$ (Eq 9.14)

Now we assume $\displaystyle h(t)=e^{\alpha t}$, where $\displaystyle {\alpha}$ is a constant

So Eq 9.14 can be expressed as

 $\displaystyle {\alpha}^2{a_2}-{\alpha}{a_1}+{a_0}=0$ (Eq 9.15)

Then we try to reduce order for Eq 9.2 and express Eq 9.2 as

 $\displaystyle {{e}^{\alpha t}}\left( {{{\bar{a}}}_{1}}{y}'+{{{\bar{a}}}_{2}}y \right)=\int{{{e}^{\alpha t}}f(t)dt}$ (Eq 9.16)

We can get

 $\displaystyle {{e}^{\alpha t}}\left[ {{{\bar{a}}}_{1}}{y}''+(\alpha {{{\bar{a}}}_{1}}+{{{\bar{a}}}_{0}}){y}'+\alpha {{{\bar{a}}}_{0}}y \right]={{e}^{\alpha t}}f(t)$ (Eq 9.17)

Therefore, we have

 \displaystyle \begin{align} & {{{\bar{a}}}_{1}}={{a}_{2}} \\ & {{{\bar{a}}}_{0}}={{a}_{1}}-\alpha {{a}_{2}} \\ & {{{\bar{a}}}_{0}}={{a}_{0}}/\alpha \\ \end{align} (Eq 9.18)

Rearrange Eq 9.16, we have

 $\displaystyle \left( {y}'+\frac{{{\bar{a}}}_{2}}{{{\bar{a}}}_{1}}y \right)=\frac{{e}^{-\alpha t}}{{{\bar{a}}}_{1}}\int{{{e}^{\alpha t}}f(t)dt}$ (Eq 9.19)

Using $\displaystyle h(t)=e^{{\beta}t}$, where $\displaystyle {\beta}=\frac{\bar{a_0}}{\bar{a_1}}$

Thus, we can get

 \displaystyle \begin{align} y(t) & =\frac{{{e}^{-\beta t}}}{{{{\bar{a}}}_{1}}}\int_{{}}^{t}{{{e}^{(\beta -\alpha )\tau }}(\int_{{}}^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \\ & =\frac{{{e}^{-\beta t}}}{{{{\bar{a}}}_{1}}}\int_{{}}^{t}{{{e}^{(\beta -\alpha )\tau }}\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau }+{{k}_{1}} \right)d\tau } \\ & =\frac{{{e}^{-\beta t}}}{{{{\bar{a}}}_{1}}}\left[ \int_{{}}^{t}{{{e}^{(\beta -\alpha )\tau }}\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau } \right)d\tau }+\int_{{}}^{t}{{{e}^{(\beta -\alpha )\tau }}{{k}_{1}}d\tau } \right] \\ & =\frac{{{e}^{-\beta t}}}{{{{\bar{a}}}_{1}}}\left[ \int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}+{{k}_{2}}+\frac{{{k}_{1}}}{\beta -\alpha }{{e}^{(\beta -\alpha )t}} \right] \\ & =\frac{{{k}_{1}}}{(\beta -\alpha ){{{\bar{a}}}_{1}}}{{e}^{-\alpha t}}+\frac{{{k}_{2}}}{{{{\bar{a}}}_{1}}}{{e}^{-\beta t}}+\frac{{{e}^{-\beta t}}}{{{{\bar{a}}}_{1}}}\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \end{align} (Eq 9.20)

Therefore, we have

 $\displaystyle y(t)={{C}_{1}}{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\beta t}}+\frac{{{e}^{-\beta t}}}{{{a}_{2}}}\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}$ (Eq 9.21)

Where $\displaystyle {C_1}$ and $\displaystyle {C_2}$ are constants.

# Contributing Members

 Team Contribution Table Problem Number Lecture(Mtg) Assigned To Solved By Typed By Signature(Author) R*5.1 Mtg 22-6 Chung Chung Chung Chung 17 Oct. 2011 at 1:30 (UTC) R*5.2 Mtg 27-1 Shin Shin Shin Shin 31 Oct. 2011 at 14:38 (UTC) R*5.3 Mtg 30-3 Ankush Ankush Ankush Ankush 1 Nov. 2011 at 18:00 (UTC) R*5.4 Mtg 31-1 YuChen YuChen YuChen YuChen 30 Oct. 2011 at 08:47 (UTC) R*5.5 Mtg 31-3 YuChen YuChen YuChen YuChen 30 Oct. 2011 at 10:01 (UTC) R5.6 Mtg 31-5 Ankush Ankush Ankush Ankush 1 Nov. 2011 at 20:00 (UTC) R*5.7 Mtg 32-1 Chung Chung Chung Chung 29 Oct. 2011 at 0:19 (UTC) R*5.8 Mtg 32-2 Shin Shin Shin Shin 2 Nov. 2011 at 15:43 (UTC) R*5.9 Mtg 32-7 Ren Ren Ren Ren 1 Nov. 2011 at 23:50 (UTC)


text:
# Rotating dielectric sphere : Jackson 6.8

## Homework Statement

I'm working at the exercise 6.8 of Jackson's Classical Electrodynamics. Here it is :
We consider a dielectric sphere of dielectric constant $$\epsilon$$ and of radius $$a$$ located at the origin. There is a uniform applied electric field $$E_0$$ in the $$x$$ direction. The sphere rotates with an angular velocity $$\omega$$ about the $$z$$ axis. We have to show that there is a magnetic field $$\textbf{H} = - \nabla \Phi_M$$ where :
$$\Phi_M(\textbf{x}) = \frac{3}{5}\epsilon_0 \left( \frac{\epsilon - \epsilon_0}{\epsilon + 2 \epsilon_0} \right) E_0 \omega \left( \frac{a}{r_<} \right)^5 x z$$
where $$r_> = \max(r, a)$$. The motion is nonrelativistic.

## The Attempt at a Solution

In every solutions I found (here is a good one : http://www-personal.umich.edu/~pran/jackson/P505/p10s.pdf" ), they suppose that the polarization of the sphere is the same as if the sphere stayed static :
$$\mathbf{P} = 3 \epsilon_0 \left( \frac{\epsilon - \epsilon_0}{\epsilon + 2 \epsilon_0} \right) E_0$$
I guess we may assume this because we suppose the motion is nonrelativistic ?
So we have a bound surface charge density
$$\sigma_{pol} = \mathbf{P} \cdot \mathbf{\hat{n}}$$
where $$\hat{n}$$ is the normal vector to the sphere.
Now comes the part I don't understand. They suppose, since the sphere is rotating, there is an effective surface current with density :
$$\mathbf{K}_M = \sigma_{pol}(\mathbf{x}) . \mathbf{v}(\mathbf{x})$$
I don't understand why we may suppose the existence of such surface current since, as I see the situation, the surface charge density stay constant because we assumed the polarization was the same as if the sphere didn't rotate...

I hope I was clear :)

Last edited by a moderator:

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Ben Niehoff


text:
# How do you solve 2x^2+7x+9=0 algebraically?

Nov 30, 2017

In fraction form:

$\setminus \rightarrow x = \setminus \frac{i \setminus \sqrt{23}}{4} - \setminus \frac{7}{4}$

Simplified form:

$\setminus \rightarrow x \setminus \approx - 1.75 \setminus \pm 1.19895788 i$

#### Explanation:

We’ll solve this two ways: by completing the square and using the quadratic formula. Here’s the first way:

$2 {x}^{2} + 7 x + 9 = 0$

First divide everything by $2$, since $a$ cannot have a coefficient:

$\setminus \rightarrow {x}^{2} + \setminus \frac{7}{2} x + \setminus \frac{9}{2} = 0$

Now, move $c$ to the RHS:

$\setminus \rightarrow {x}^{2} + \setminus \frac{7}{2} x = - \setminus \frac{9}{2}$

Now add ${\left(\setminus \frac{b}{2}\right)}^{2}$ to both sides. Here, $b = 7$.

$\setminus \rightarrow {x}^{2} + \setminus \frac{7}{2} x + \setminus \frac{49}{16} = \setminus \frac{49}{16} - \setminus \frac{9}{2}$

Simplify the RHS:

$\setminus \rightarrow {x}^{2} + \setminus \frac{7}{2} x + \setminus \frac{49}{16} = - \setminus \frac{23}{16}$

Factor the LHS into ${\left(x + \setminus \frac{b}{2}\right)}^{2}$:

$\setminus \rightarrow {\left(x + \setminus \frac{\setminus \frac{7}{2}}{2}\right)}^{2} = - \setminus \frac{23}{16}$

$\setminus \rightarrow {\left(x + \setminus \frac{7}{4}\right)}^{2} = - \setminus \frac{23}{16}$

Take the square root of both sides:

$\setminus \rightarrow x + \setminus \frac{7}{4} = \setminus \sqrt{- \setminus \frac{23}{16}}$

Isolate $x$:

$\setminus \rightarrow x = \setminus \sqrt{- \setminus \frac{23}{16}} - \setminus \frac{7}{4}$

$\setminus \rightarrow x = i \setminus \sqrt{\setminus \frac{23}{16}} - \setminus \frac{7}{4}$

$\setminus \rightarrow x = i \setminus \frac{\setminus \sqrt{23}}{4} - \setminus \frac{7}{4}$

$\setminus \rightarrow x \setminus \approx - 1.75 \setminus \pm 1.19895788 i$

Here’s the second way, using the quadratic formula:

$2 {x}^{2} + 7 x + 9 = 0$

First, we need to identify $a$, $b$, and $c$:

$a = 2$

$b = 7$

$c = 9$

Now, plug them into the formula:

$x = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\setminus \rightarrow x = \setminus \frac{- 7 \setminus \pm \setminus \sqrt{{7}^{2} - 4 \left(2\right) \left(9\right)}}{2 \left(2\right)}$

$\setminus \rightarrow x = \setminus \frac{- 7 \setminus \pm \setminus \sqrt{49 - 72}}{4}$

$\setminus \rightarrow x = \setminus \frac{- 7 \setminus \pm \setminus \sqrt{- 23}}{4}$

$\setminus \rightarrow x = \setminus \frac{- 7 \setminus \pm i \setminus \sqrt{23}}{4}$

Rearranging yields:

$\setminus \rightarrow x = \setminus \frac{i \setminus \sqrt{23}}{4} - \setminus \frac{7}{4}$

$\setminus \rightarrow x \setminus \approx - 1.75 \setminus \pm 1.19895788 i$

It’s the same answer we got by completing the square, so we know the answer is correct.


text:
# Math Help - Compute

1. ## Compute

Compute: 3/8 + 4/5

I don't understand this questions, is from my placement test, I solved it but it's not a choice of the test. Is it trying to fool me? or am I fooling myself.

My answer: 47/40 or $1\frac{7}{40}$

2. ## Re: Compute

Originally Posted by vaironxxrd
Compute: 3/8 + 4/5

I don't understand this questions, is from my placement test, I solved it but it's not a choice of the test. Is it trying to fool me? or am I fooling myself.

My answer: 47/40 or $1\frac{7}{40}$
Your work is correct.

3. ## Re: Compute

Originally Posted by vaironxxrd
Compute: 3/8 + 4/5
I don't understand this questions, is from my placement test, I solved it but it's not a choice of the test. Is it trying to fool me? or am I fooling myself.
My answer: 47/40 or $1\frac{7}{40}$
CHECK it YOURSELF (use a calculator!):
3/8 + 4/5 = 1.175
47/40 = 1.175

4. ## Re: Compute

Originally Posted by Plato
Your work is correct.
Originally Posted by Wilmer
CHECK it YOURSELF (use a calculator!):
3/8 + 4/5 = 1.175
47/40 = 1.175
Therefore, the test answers are incorrect? Why is from the compass sample questions...

5. ## Re: Compute

Originally Posted by vaironxxrd
Therefore, the test answers are incorrect? Why is from the compass sample questions...
Ask THEM, not us...


text:
# Line integral

1. Nov 7, 2011

### stratusfactio

1. The problem statement, all variables and given/known data

F=<xy, x-y> and C is the triangle joining (1,0), (0,1) and (-1,0) in the clockwise direction.

2. Relevant equations

How do I have the second part of this curve?

3. The attempt at a solution
Apparently, there should only be a sum of two integrals. So I got that curve one could be
$$C_1: x=t, y=0, -1≤ t ≤ 1$$

But I have no idea how to find $$C_2$$. There's no other way, that I see, that you can take such that you keep x constant and let y vary.

2. Nov 8, 2011

### tjackson3

The reason it should only be a sum of two integrals is because the line integral is just zero on the curve you have. You're only interested in the sum of the integral over $C_2$ and $C_3$. How might you parametrize those? For example, if you wanted to find a line in x-y coordinates to go from (1,0) to (0,1), you would find its slope (-1) and its y-intercept (1). The line would be y = 1-x. So if you let x = t, then what would y be and what would t vary between?


text:
05/07/18

#### Find K such that the system of equations has 1 unique solution.

-2x -7y +1z=46x -5y -3z=14kx -2y -z=5 k=? I need help understanding how to solve this problem. Steps and answers would be very helpful. Thank you!

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seven hundred grams of a radioactive material decays according to the formula A=700(2/3)^2, where t is measured in years. Find the amount present in 10 years. Round to the nearest one tenth of a gram  more

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15/(x^2-5x) is the rational expression. I got as far as 15/(x(x-5)). I don't know where to go from here.

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2/x-3 + 1/4 = 2

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(0.008)∧−2/3

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