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linear ordinary differential equation

Definition/Summary
An nth-order linear ordinary differential equation (ODE) is a differential equation of the form $$\sum_{i=0}^n a_i(x)y^{(i)}(x)\ =\ b(x)$$ where $y^{(i)}(x)$ denotes the ith derivative of y with respect to x.  The difference between any two solutions is a solution of the homogeneous part: $$\sum_{i=0}^n a_i(x)y^{(i)}(x)\ =\ 0$$  This equation has n independent solutions, and every solution is a linear combination of them.  So the general solution of a linear ODE is the sum of one particular solution of the whole equation plus a linear combination of n independent solutions of the homogeneous part (in other words: a particular solution plus any homogeneous solution).

Equations

Scientists
Joseph Louis Lagrange (1736-1813)

Recent forum threads on linear ordinary differential equation

Breakdown
Mathematics 					> Calculus/Analysis 					>> Differential Equations

See Also
recurrence relation

Images

Extended explanation
Homogeneous linear ODE:  A homogeneous linear ODE is a linear ODE in which $b(x)=0.$  Homogeneous solutions:  These are linear combinations of solutions of the form $e^{\lambda x}$ or $x^ie^{\lambda x}$.  Any solution of the form $e^{\lambda x}$ can be found by the method shown in the following example.  For the general method of solution, see the method of characteristic polynomial, at the foot of this page.  Example of homogeneous solution:  Find the general solution to $y'' - 2y' -8y = \sin{x}+ \cos{x}$  First we must find the general homogeneous solution, $y_h$.  Let's guess that $y_h$ is in the form $y_h=e^{\lambda x}$ - this seems like  a good guess, doesn't it? So, plugging in, we have $$\left(\lambda^2 - 2\lambda -8\right) e^{\lambda x} = 0$$ Now, since $e^{\lambda x}$ is never $0$, we can solve the above equation as a quadratic (the characteristic quadratic), and we find that either $\lambda=4$ or $\lambda=-2$. Therefore the general solution is $$y_h = C_1 e^{4x} + C_2 e^{-2x}$$ where $C_1$ and $C_2$ are arbitrary constants.  Particular solutions:  We only need one particular solution.  Finding one is generally a combination of common-sense and guesswork, based on the nature of the polynomial $b(x)$.  Let's see how to do it with the above example, in which $b(x)\ =\ \sin{x}+ \cos{x}$, using the method of undetermined coefficients …  There is also the more general variation of parameters method, but this is not usually needed in examination questions.   Example of Particular Solution by the Method of Undetermined Coefficients:  Let's guess a particular solution in the form $y_p\ =\ A\sin{x} + B\cos{x}$.  The method of undetermined coefficients is to solve for $A$ and $B$ by plugging $y\ =\ y_p$ into the original equation, giving:  $$\sin{x}(-9A\ +\ 2B)\ +\ \cos{x}(-2A\ -\ 9B)\ =\ \sin{x}\ +\ \cos{x}$$  which, dealing with the coefficients of $\sin{x}$ and $\cos{x}$ separately, gives:  $$A\ =\ -11/85\ \ \ \ B = -7/85$$  and so $y_p\ =\ -(11\sin{x}\ +\ 7\cos{x})/85$ is a solution to the original equation.  Using this as the particular solution, the general solution of the original equation is:  $$y= C_1 e^{4x} + C_2 e^{-2x} - (11\sin{x}\ +\ 7\cos{x})/85$$  Homogeneous solution by characteristic polynomial:  This is not the same as the characteristic polynomial of a matrix or matroid  In a linear differential equation, the derivative may be replaced by an operator, D, giving a polynomial equation in D:  $$\sum_{n\,=\,0}^m\,a_n\,\frac{d^ny}{dx^n}\ =\ 0\ \mapsto\ \left(\sum_{n\,=\,0}^m\,a_n\,D^n\right)y\ =\ 0$$  If this polynomial has distinct (different) roots $\lambda_1,\dots,\lambda_m$:  $$\prod_{n\,=\,1}^m(D\,-\,\lambda_n)\ =\ 0$$  then the general solution is a linear combination of the solutions of each of the equations:  $$\left(D\,-\,\lambda_n\right)y\ =\ 0$$  which are the same as $$\frac{dy}{dx}\ =\ \lambda_n\,y$$  and so the general solution is of the form:  $$y\ =\ \sum_{n\,=\,1}^m\,C_n\,e^{\lambda_nx}$$  For a pair of complex roots (they always come in conjugate pairs) $p\ \pm\ iq$ or $r\,e^{\pm is}$, a pair of $C_ne^{r_nx}$ may be replaced by $$e^{px}(A\,cos(qx)\,+\,iB\,sin(qx))$$ or $$r^k(A\,cos(sk)\,+\,iB\,sin(sk))$$  However, if the polynomial has some repeated roots:  $$\prod_{p\,=\,1}^q\prod_{n\,=\,1}^p(D\,-\,\lambda_n)^p\,y\ =\ 0$$  then the general solution is of the form:   $$y\ =\ \sum_{p\,=\,1}^q\sum_{n\,=\,1}^pC_{n,p}\,x^{p-1}\,e^{\lambda_nx}$$

Commentary

paulojomaje @ 03:27 AM Dec18-10
too difficult

tiny-tim @ 11:47 AM Jan16-09
foxjwill's original draft of this page, which got lost in the preview process, is now restored, with amendments to Definition and other minor editing. Method of characteristic polynomial added.

foxjwill @ 09:57 PM Jun19-08
yes, it is. Actually, when I wrote it originally, it said that I wasn't able to post it so, until today, I actually hadn't known it was here.

mathwonk @ 01:12 PM May10-08
the simplest linear homogeneous constant coeff equation has form (D-c)f = 0, where D denotes differentiation, and the most complicated one is merely a formal product of these.  Since such products commute with each other, it suffices to solve this one and its iterates, of form (D-c)^n f = 0.  clearly, the solution of Df = 0 is a constant, and the solution of (D-c)f = 0 is a constant multiple of e^(ct).  Moreover a linear combination of solutions is again a solution, so at least when all factors (D-c) occur simply in our operator, the solutions are linear combinations of the functions e^(ct).  as an exercise, check then that solutions of (D-c)^2 f = 0, are exactly solutions of (D-c)f = constant times e^(ct), or linear combinations of te^(ct) and e^(ct).

FunkyDwarf @ 07:30 AM May10-08
Err is this meant to be incomplete or...
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