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Let's say I have the DE:

$$ (x^2 - 2x)y'' - (x^2 - 2)y' + (2x - 2)y = 0 $$

And I have one possible solution to the DE:

$$ y_1(x) = e^x $$

How would I go about solving this? I could solve the actual DE, but then what is the point of supplying a possible solution? Where does the solution $y_1$ come into play?

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I can't tell if you're asking this in total generality (ie, $f(x)$, $g(x)$, $h(x)$ are arbitrary) or if you are abstracting the specifics to convey the "gist" of your problem. One thing that comes to mind is that if you know the coefficients of an equation $y'' + p(x) y' + q(x) y = 0$ (your equation has this form, at least near values of $x$ where $f(x)$ is nonzero) you can, via "Abel's formula", write down the Wronskian of any two solutions to the equation. With this and one solution to the DE, you have a first order DE you can try to solve for another solution to the original equation. –  leslie townes Feb 5 '12 at 20:58
    
In you example, by direct substitution, you are forced to admit that f(x)-g(x)+h(x)=0$ and the only possible solution is that these are constants. –  Jon Feb 5 '12 at 20:58
    
@leslietownes: I've added specifics to my question, if that helps. –  MaxMackie Feb 5 '12 at 21:10

2 Answers 2

up vote 4 down vote accepted

You can proceed using Abel's integration identity. In general for differential equations of the form $$ \sum\limits_{k=0}^n a_k(x)y^{(n-k)}(x)=0 $$ we can consider its solutions $y_1(x),\ldots,y_n(x)$ and define so called Wronskian $$ W(y_1,\ldots,y_n)(x)= \begin{pmatrix} y_1(x)&&y_2(x)&&\ldots&&y_n(x)\\ y'_1(x)&&y'_2(x)&&\ldots&&y'_n(x)\\ \ldots&&\ldots&&\ldots&&\ldots\\ y'_1(x)&&y'_2(x)&&\ldots&&y'_n(x)\\ \end{pmatrix} $$ Then we have the following identity $$ \det W(x)=\det W(x_0) e^{-\int\limits_{x_0}^x \frac{a_1(t)}{a_0(t)}dt} $$ In particular for your problem we have the following differential equation $$ \begin{vmatrix} y_1(x)&&y_2(x)\\ y'_1(x)&&y'_2(x) \end{vmatrix}=C e^{-\int\frac{-(x^2-2)}{x^2-2x}dx} $$ with $y_1(x)=e^x$. Which reduces to $$ y'_2(x)e^x-y_2(x)e^x=C e^{\int\frac{x^2-2}{x^2-2x}dx}=C(2x-x^2)e^x $$ After division by $e^{2x}$ we get $$ \frac{y'_2(x)e^x-y_2(x)e^x}{e^{2x}}=C(2x-x^2)e^{-x} $$ which is equivalent to $$ \left(\frac{y_2(x)}{e^x}\right)'=C(2x-x^2)e^{-x} $$ It is remains to integrate $$ \frac{y_2(x)}{e^x}=Cx^2 e^{-x}+D $$ and write down the answer $$ y_2(x)=Cx^2+D e^{x} $$ In fact this is a general solution of original equation.

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$$ (x^2 - 2x)y'' - (x^2 - 2)y' + (2x - 2)y = 0 $$

Let's first notice that $c y_1(x)= c e^x$ is also a solution.

To find other solutions let's suppose that $c$ depends of $x$ (this method is named 'variation of constants') :

If $y(x)= c(x) e^x$ then your O.D.E. becomes : $$ (x^2 - 2x)(c''+c'+c'+c)e^x - (x^2 - 2)(c'+c)e^x + (2x - 2)ce^x = 0 $$ $$ (x^2 - 2x)(c''+2c'+c) - (x^2 - 2)(c'+c) + (2x - 2)c = 0 $$ Of course the $c$ terms disappear and we get :

$$ (x^2 - 2x)(c''+2c') - (x^2 - 2)c' = 0 $$ Let's set $d(x)=c'(x)$ then : $$ (x^2 - 2x)d' = (x^2 - 2)d-(x^2 - 2x)2d $$ $$ (x^2 - 2x)d' = (-x^2 +4x- 2)d $$

$$ \frac{d'}d = \frac{-x^2 +4x- 2}{x^2 - 2x} $$ I'll let search the integral at the right, the answer should be ($C_0$, $C_1$, $C_2$ are constants) : $$ \ln(d)=\ln(x^2-2x)-x+C_0 $$

$$ d=(x^2-2x)e^{-x}C_1 $$
but $c'=d$ so that $$ c=C_2+C_1\int (x^2-2x)e^{-x} dx $$

$$ c=C_2-C_1x^2e^{-x} $$ And we got the wished general solution : $$ y(x)=c(x)e^x=C_2e^x-C_1x^2 $$

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