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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.2012-02-05
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    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.2012-02-05
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    @leslietownes: I've added specifics to my question, if that helps.2012-02-05

2 Answers 2

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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 : $$ y(x)=c(x)e^x=C_2e^x-C_1x^2 $$