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## 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

[tex]a_0(x)y + a_1(x)y' + \cdots + a_{n-1}(x)y^{(n-1)} + y^{(n)} = b(x)\qquad\qquad\qquad(1)[/tex]

Now, recall that the product rule for derivatives of order [tex]n[/tex] is

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

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

[tex]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)[/tex]

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

[tex]

\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)[/tex]

\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)[/tex]

Rearranging [tex](3)[/tex], we have

[tex]

\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)[/tex]

\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)[/tex]

Solving each equation in [tex](4)[/tex], we get

[tex]u^{(k)} = e^{\binom{n}{n-k} {\int \frac{a_{n-k}}{a_{n-(k+1)}}dx }}[/tex].

Setting [tex]k=0[/tex],

[tex]u(x) = e^{\int \frac{a_n(x)}{a_{n-1}(x)}dx}\qquad\qquad\qquad(5)[/tex]

Now, going back to [tex](1)[/tex] and [tex](2)[/tex], since, by definition,

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

we can plug in [tex](5)[/tex],

[tex]\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)[/tex]

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

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