# Solving general linear ODE's

## Homework Statement

Is there a standard method for finding the general solution to any given linear homogeneous ODE?

I tried working it out myself. Here's what I tried. I'm almost positive it's wrong because it doesn't take into account most of the coefficients. But maybe I'm on the right track?

## The Attempt at a Solution

$$a_0(x)y + a_1(x)y' + \cdots + a_{n-1}(x)y^{(n-1)} + y^{(n)} = b(x)\qquad\qquad\qquad(1)$$​

Now, recall that the product rule for derivatives of order $$n$$ is

$$\frac{d^n}{dx^n}[f(x)g(x)] = {\sum^n_{k=0} \binom{n}{k} f^{(k)}(x)g^{(n-k)}}(x).$$​

We want to find some function $$u(x)$$ that, when multiplied to both sides of $$(1)$$, will put the left side in the form of the $$n$$th degree product rule. In other words,

$$u(x)a_0(x)y + u(x)a_1(x)y' + \cdots + u(x)y^{(n)} = \frac{d^n}{dx^n}[u(x)y]\qquad\qquad\qquad(2)$$​

After expanding $$(2)$$, we can set each term of the left side equal to the corresponding term on the right (e.g. $$u(x)a_2(x) y' = u^{(n-1)}(x)y'$$ ) and, after canceling out the all the $$y$$'s, form a system. Also note that since $$u(x)y^{(n)}$$ appears on both sides of $$(2)$$ it cancels out. From here until I say otherwise, all functions will be referenced, for convenience, without the "$$(x)$$" appended to it.

$$\left \{ \begin{array}{lcl} u^{(n)} &=& u a_0\\ u^{(n-1)} &=& \dbinom{n}{1} u a_1\\ & \vdots\\ u'' &=& \dbinom{n}{n-2} u a_{n-2}\\ u' &=& \dbinom{n}{n-1} u a_{n-1}\\ \end{array} \right . \qquad\qquad\qquad(3)$$​

Rearranging $$(3)$$, we have

$$\left \{ \begin{array}{lcl} \dfrac{u^{n}}{u^{(n-1)}} &=& \dfrac{a_1}{a_0} \\ \\ \dfrac{u^{(n-1)}}{u^{(n-2)}} &=& \dbinom{n}{1} \dfrac{a_2}{a_1}\\ & \vdots\\ \dfrac{u'''}{u''} &=& \dbinom{n}{n-2} \dfrac{a_{n-2}}{a_{n-3}}\\ \\ \dfrac{u''}{u'} &=& \dbinom{n}{n-1} \dfrac{a_{n-1}}{a_{n-2}}\\ \end{array} \right . \qquad\qquad\qquad(4)$$​

Solving each equation in $$(4)$$, we get

$$u^{(k)} = e^{\binom{n}{n-k} {\int \frac{a_{n-k}}{a_{n-(k+1)}}dx }}$$.​

Setting $$k=0$$,

$$u(x) = e^{\int \frac{a_n(x)}{a_{n-1}(x)}dx}\qquad\qquad\qquad(5)$$​

Now, going back to $$(1)$$ and $$(2)$$, since, by definition,

$$\frac{d^n}{dx^n}[u(x)y] = u(x)b(x),$$​

we can plug in $$(5)$$,
$$\frac{d^n}{dx^n}\left[ e^{\int \frac{a_n(x)}{a_{n-1}(x)}dx} y \right] = e^{\int \frac{a_n(x)}{a_{n-1}(x)}dx}b(x)\qquad\qquad\qquad(6)$$​

and then antidifferentiate, which, at the moment, I am feeling too lazy to actually do.

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rock.freak667
Homework Helper
Are there not various methods to solve first order ODE's with constant coefficients such as variables are separable,integrating factor and such?

Are there not various methods to solve first order ODE's with constant coefficients such as variables are separable,integrating factor and such?
Yes, but I was wondering about methods to solve general degree ODE's with variable coefficients.

You have introduced one function $$u(x)$$ and you also have n equations in equation (3).

In general your solution for $$k=0$$, i.e. equation (5), does not satisfy all members of equation (3).
Thus there is no $$u(x)$$ in general.

HallsofIvy
Homework Helper
Series solutions (including Frobenius' method) are the most general methods of solving linear equations.

You have introduced one function $$u(x)$$ and you also have n equations in equation (3).

In general your solution for $$k=0$$, i.e. equation (5), does not satisfy all members of equation (3).
Thus there is no $$u(x)$$ in general.
Yeah. I had a feeling it was something like that.

Series solutions (including Frobenius' method) are the most general methods of solving linear equations.
Hm. Darn. I don't like using series solutions. They're so cumbersome. Ah, well.

Also, I'm assuming the laplace and fourier transforms, in general, can't be used to solve linear ODE's with variable coefficients? I know that there are a few specific cases where they work, but, yeah.