- #1
foxjwill
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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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