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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
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 nth 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.
<br />
\left \{<br />
\begin{array}{lcl}<br />
u^{(n)} &=& u a_0\\<br />
u^{(n-1)} &=& \dbinom{n}{1} u a_1\\<br />
& \vdots\\<br />
u'' &=& \dbinom{n}{n-2} u a_{n-2}\\<br />
u' &=& \dbinom{n}{n-1} u a_{n-1}\\<br />
\end{array}<br />
\right .<br />
\qquad\qquad\qquad(3)
Rearranging (3), we have
<br />
\left \{<br />
\begin{array}{lcl}<br />
\dfrac{u^{n}}{u^{(n-1)}} &=& \dfrac{a_1}{a_0} \\<br />
\\<br />
\dfrac{u^{(n-1)}}{u^{(n-2)}} &=& \dbinom{n}{1} \dfrac{a_2}{a_1}\\<br />
& \vdots\\<br />
\dfrac{u'''}{u''} &=& \dbinom{n}{n-2} \dfrac{a_{n-2}}{a_{n-3}}\\<br />
\\<br />
\dfrac{u''}{u'} &=& \dbinom{n}{n-1} \dfrac{a_{n-1}}{a_{n-2}}\\<br />
\end{array}<br />
\right .<br />
\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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