DE: $x^3y'''+4x^2y''-5xy'-15y=x^4$
Making the substitution $y=x^m$ to develop the characteristic equation $m^3+m^2-7m-15=0$ gives roots of $3,-2{\pm}i$. From this homogeneous general solution is
$c_1e^{3x}+e^{-2x}(c_2\cos (x)+c_3\sin(x))$.
Solving using method of undetermined coefficients, Kreyszig suggests if $r(x)$ is in $x^n$ form to assume a solution in the form of $K_nx^n+K_{n-1}x^{n-1}+...+K_0$. Doing this ends with a particular solution of $x^4/37$.
I have access to the solution, which is $y=(1/x^2)(c_2\cos(\ln x)+c_3\sin(\ln x))+c_1x^3 + x^4/37$.
Why was the substitution $x=\ln x$ done in the homogeneous portion of the solution?
Thank you.
[I am using Advanced Engineering Mathematics by Kreyszig for reference]
Edit: I actually just read that method of undetermined coefficients is only for constant coefficients, in this case $x$ isn't constant so I am not sure why it worked out for the particular solution, maybe I got lucky.