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Consider the second order ODE where

$ (k-x)^2 y''+6(k-x)y'+12y=F(x) $

where $k$ is some constant. I want to compute the real valued general solution.

progress: guess $(k-x)^{m}$ to be the solution and I find m to be of the form $m=a+bi$. Applying variation of parameters, it takes me into a long algebra mess.

Any trick/ways to do this quickly. ($y(0)$ and $y'(0)$ are given.)

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Is the ODE supposed to be $(k - x)^2 y'' + 6(k-x) y' + 12y = F(x)$? If so, you should be guessing $(k-x)^m$ as the form of the solution, since $y$ is the function to be solved. If the equation you've written is supposed to be correct, then the equation is nonlinear and variation of parameters does not apply. –  Christopher A. Wong Jan 23 '13 at 19:49
    
thanks, I meant what you said above and I have edited my question now. –  felasfa Jan 23 '13 at 19:56
    
What is $F(x)$? –  Christopher A. Wong Jan 23 '13 at 20:01
    
$F(x)=(1-c)(x-k)^2+6(x-k)-24$. –  felasfa Jan 24 '13 at 22:20
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2 Answers 2

If that is not cheating, you could try a computer algebra package, like maxima. At least to help with the "algebra mess" you mention.

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First, solve the homogeneous differential equation which is of Euler differential equation type and has a solution

$$ y \left( x \right) ={c_1}\, \left( x-k \right) ^{3}+{c_2}\, \left( x-k \right) ^{4}.$$

So, the fundamental set of solutions is given by $ \left\{ \left( x-k \right) ^{3},\left( x-k \right) ^{4} \right\}. $

You can see that the solutions are linearly independent, since the Wronskian is $-(k-x)^6$. Now just advance with the method of variation of parameters to find the particular solution.

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Thanks Mhenni,I guessed the solution to be of the form $(x-k)^{m}$ and when I solved for the char.roots, I did it wrong and got $m$ to be a complex number and went to a wrong route when I used the VOP. –  felasfa Jan 24 '13 at 22:23
    
@abiyo: You are welcome. –  Mhenni Benghorbal Jan 26 '13 at 0:12
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