# Please Check The Solution of my ODE

hey guys just need to know if my differential equation is right

the question is

b)$$\frac{dy}{dx} = x + 6y.$$

the question is :Find the general solutions to the following differential equations. Sketch at least 4 solution curves for each.

the answer that i have found is $$y=\sqrt{\frac{x^2}{6}}$$

please let me know if thats the right answer

thanks

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– nik Feb 5 '12 at 13:22

## 4 Answers

Your answer seems to be wrong. ($\sqrt{x^2}$ makes little sense as it is)

$$y = \frac{(Ce^{6x})}{36} - \frac{x}{6} - \frac{1}{36}$$ You can verify this: $$\frac{dy}{dx} = \frac{(Ce^{6x})}{6} - \frac{1}{6}\cdots\cdots\cdots(A)$$ Now, $$6y = \frac{(Ce^{6x})}{6} - \frac{x}{1} - \frac{1}{6}$$ $$x+6y = \frac{(Ce^{6x})}{6} +x- \frac{x}{1} - \frac{1}{6}$$ $$x+6y = \frac{(Ce^{6x})}{36} -\frac{1}{6} \cdots\cdots\cdots (B)$$ Compare (A) and (B) By putting various values of C, you can plot your graphs. What method did you use to solve the ODE?

One way of solving it (Laplace Transform): $$\frac{dy}{dt} = t + 6y.$$ (Just changed variable from x to t) $$L(\frac{dy}{dt}) = L(t) + L(6y)$$ $$sL(y) - f'(0) = \frac{1}{s^2} + 6 L(y)$$ $$(s-6)L(y) = f'(0) + \frac{1}{s^2}$$ $$L(y) = \frac{f'(0) + \frac{1}{s^2}}{s-6}$$ $$y = L^{-1}(\frac{f'(0) + \frac{1}{s^2}}{s-6} )$$ $$y = f'(0)e^{-6t} + \frac{1}{s^2(s-6)}$$ Then you can use partial fractions and solve for the other two terms

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i used the separable method that was taught in my class. can you please tell me how to correct it? thanks means a lot – Rohit Pradeep Feb 5 '12 at 12:29
Since this is a homework question, I would highly recommend you post your solution so we can debug it. Giving an answer will not help you get into the IIT :P – Inquest Feb 5 '12 at 12:32
its not a homework solution im geting practice for my test  the answer is y=sqrt(x^2/6) i dnt know if that is right – Rohit Pradeep Feb 5 '12 at 12:44
If you have the background for Laplace transform, I have edited my answer with a solution using that. – Inquest Feb 5 '12 at 12:48
can you please tell me how to do it in the separable method? please i will understand it better – Rohit Pradeep Feb 5 '12 at 12:51
show 2 more comments

This is First Order Linear Differential Equation so general solution is given by :

$$y=\frac{\int u(x) \cdot x \,dx+C}{u(x)} ,\text {where}~~ u(x)=e^{-6\int \,dx}$$

Therefore solution is :

$$y=\frac{\int e^{-6x} \cdot x \,dx+C}{e^{-6x}}$$

Integral from the last equation can be solved using Integration by parts .

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One of the first methods you learn to solve an equation like this is integrating factors. First let's subtract $6y$ from both sides to get.

$y'-6y=x$

The proper integrating factor here is $e^{-6x}$. Multiplying both sides of the equation by this gets us

$e^{-6x}y'-6e^{-6x}y=xe^{-6x}$

The left side can be rewritten to obtain

$(e^{-6x}y)'=xe^{-6x}$

Can you take it from there?

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cnt i do it from the separable method? – Rohit Pradeep Feb 5 '12 at 21:35
@Rohit Not sure how Kannappan's solution works, but the equation as it stands is not separable.  I don't see any way to get this equation into the form $\int f(x)dx=\int g(y)dy$ – Mike Feb 6 '12 at 12:30

Here we'll use the variable seperable method:

Set $x+6y =t \implies 1+6\dfrac{\mathrm d y}{\mathrm {dx}}=\dfrac{\mathrm d t}{\mathrm{dx}}$. So, the equation transforms to the following:

$$\begin{align*}1+6t = \dfrac{\mathrm dt}{\mathrm{dx}}& \implies \dfrac{\mathrm dt}{1+6t}=\mathrm{dx} \end{align*}$$

Now, integrate on both sides to see what you get.

If this does not get you to the result, I'll help you a bit more!

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ok this is what i had with the same method -6y dy/dx =x    integrate both sides i get   -6y^2/2=x^2  which is =-3y^2/x^2/2 im kinda lost after this – Rohit Pradeep Feb 5 '12 at 13:18
@Rohit How did you get $-6y \cdot \dfrac{dy}{dx}=x$? – user21436 Feb 5 '12 at 13:24
i tried to use the separable method, so i separated the y from the x. – Rohit Pradeep Feb 6 '12 at 6:37
@Rohit Add a sketch of steps in your question, if you would want to know where you are mistaken. But otherwise, isn't my solution clear? – user21436 Feb 6 '12 at 6:40
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