How do you solve the ODE y′′−4y′+4y=(x+1)e2x using variation of parameter?

Answer 1

Hi,

First you solve the homogenous equation y"-4y'-4=0, assuming y=e^kx gives k²-4k-4=0, the roots are 2,2 ie repeated real roots so the linearly independent solutions are y₁=e^2x and y₂=xe^2x The Wronskian W(x) is y₁*y₂^' -y₂*y₁^'= e^4x =0 so y₁ and y₂ are indeed LI. Now if we define f(x)=e^2x *(1+x) which is the RHS of the original ODE then a particular solution is y(x)=u₁(x)*y₁(x)+u₂(x)*y₂(x) Where u₁ and u₂ are given by

u₁(x)=-∫(y₂(t)*f(t)/W(t))dt and u₂(x)=∫(y₁(t)*f(t)/W(t))dt

You just have to do substitutions and integrate all of the functions are simple. Let me know if yo have any trouble.

Regards

Jim