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solve the equation $(x-1)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+y=(x-1)^2$ given that $x$ and $e^x$ are the particular integrals of the equation without the right hand member.

What is the right hand member here?


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I think the problem is saying those are the solutions of the homogeneous equation $(x-1)y''-xy'+y=0$ – Mike Feb 20 '13 at 10:22

3 Answers 3

It means that $x$ and $e^x$ are solutions of the homogeneous associated equation:

$$(x-1)\frac{d^2y}{dx^2}-x\frac{dy}{dx}+y=0 $$

The right hand side is $(x-1)^2$

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Divide both sides of this equation by $x-1$:

$$y''-\frac{x}{x-1} y' + \frac{1}{x-1} y = x-1$$

The right-hand side is $x-1$.

Multiply through by $e^{-x}/(x-1)$. This is an integrating factor which puts the equation in a Sturm-Liouville form:

$$\frac{d}{dx} \left [\frac{e^{-x}}{x-1} y'\right] + \frac{e^{-x}}{(x-1)^2} y = e^{-x}$$

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is there a step wise process to solve this. – Rajesh K Singh Feb 20 '13 at 10:42
Actually, not really in this form. I thought it more useful to see the solutions of the homogeneous equation more clearly. In general, SL form is more useful for solving when the equation is homogeoneous and is used to find eigenvalues of the differential operator defined by the left hand side. See if you are interested. – Ron Gordon Feb 20 '13 at 10:47
@ rlgordonma: thanks !! – Rajesh K Singh Feb 20 '13 at 11:01

Just solve it by variation of parameters.

$y_1=x$ , $y_2=e^x$

The Wronskian of these two functions is $\begin{vmatrix}x&e^x\\1&e^x\end{vmatrix}=(x-1)e^x$

$\therefore y=C_1x+C_2e^x-x\int\dfrac{1}{(x-1)e^x}e^x(x-1)~dx+e^x\int\dfrac{1}{(x-1)e^x}x(x-1)~dx=C_1x+C_2e^x-x\int~dx+e^x\int xe^{-x}~dx=C_1x+C_2e^x-x^2-e^x\int x~d(e^{-x})=C_1x+C_2e^x-x^2-x+e^x\int e^{-x}~dx=C_1x+C_2e^x-x^2-1$

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