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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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    @leslietownes: I've added specifics to $m$y 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 $