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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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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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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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