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I'm preparing for my final and came across a problem in the practice set which our professor didn't post the answers to... I tried to solve this question but failed to integrate $u_1'$ and $u_2'$. Here is the ODE: $$y''+y'-2y=\ln(x)$$

Could someone please help me with this? Thanks so much!!!

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It seems the integrals aren't elementary... Here's WA's solution. –  David Mitra May 8 '12 at 22:42
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First find the solution to the corresponding homogeneous equation $y\prime\prime+y\prime-2y=0$ by using characteristic equation $r^2+r-2=0$ –  Kirthi Raman May 8 '12 at 22:59
    
Thanks very much! –  hello.world May 9 '12 at 1:38

1 Answer 1

Having found two solutions of homogeneous equation, $u_1(x)=e^{-2x}$ and $u_2(x)=e^{x}$, and computed the Wronskian $W=3e^{-x}$, we end up with general solution $u=Au_1+Bu_2$ where $$A=-\int \frac{u_2}{W} \ln(x)\,dx = -\frac13 \int e^{2x}\ln x\,dx$$ $$B=\int \frac{u_1}{W} \ln(x)\,dx = -\frac13 \int e^{-2x}\ln x\,dx$$ All we can do with these is integrate by parts (differentiating $\ln$), which expresses both integrals in terms of Exponential integral $\mathrm{Ei}$. I doubt that this was professor's intent, though.

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