Quickest way to solve PDE with variation of parameters I have an exam question to solve the following PDE:
$$
  x^2y^{''}-4xy^{'}+6y=21x^{-4}
$$
This problem is supposed to be solved by the variation of parameter method(Wronskian). For finding the solution of the homogenous part I used the Euler cauchy method, but it turns out it gets very computationally intensive. What is the quickest way (and less prone to mistakes) to do this? (With the variation of parameters)
Assuming that the problem is $$x^2y''-4xy'+6y=21x^{-4}$$ For the homogeneous equation, assume $y=x^a$ and replace. This let you with $$(a-3) (a-2) x^a=0$$ So, you have $$y=c_1x^2+c_2x^3$$
After you find the solutions to the corresponding homogeneous ODE, you can find $y_p$ by variation of parameters, then the general solution to the original ODE is the sum of $y_p$ and a general linear combination of the $x^2$ and $x^3$.

Alternatively, you assume that the general solutions is something like $y=v*x^2$ ($x^2$ is a solution to the corresponding homogeneous ode, and $v$ is a function of $x$). Then, $y'=v'*x^2+2vx$ and $y''=v''*x^2+4v'x+2v$.
Substituting $y$ and its derivatives in the original ode, you end up with
$v"=21x^{-8}$

Integrate twice to get $v$, then multiply by $x^2$ to get $y$.

Hope this helps