Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

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

share|improve this question
    
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 $(x-1)^2$

share|improve this answer

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:

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

share|improve this answer
    
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 en.wikipedia.org/wiki/Sturm%E2%80%93Liouville_theory 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 $\begin{vmatrix}x&e^x\\1&e^x\end{vmatrix}=(x-1)e^x$

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.