Solving x²y'' - 3xy' + 13y = 4 + 3x with substitution x = e^t

[hola]

1. Use substitution x = e^t to transfer equation into one with constant coefficients and solve:
x^2 y'' -3xy' + 13y = 4 + 3x
My work:
Okay, we have
e^2t y'' - 3e^t y' + 13y = 4 + 3e^t.

Now I am absolutely stuck. No idea how to solve it. Help!

 

[HallsofIvy]

You didn't complete the substitution! Your y' and y" are still with respect to x, not t.

Use the chain rule: dy/dx= dy/dt dt/dx. Since x= e^t, t= ln x and
dt/dx= 1/x. That is, dy/dx= (1/x)(dy/dt).
d²y/dx²= d((1/x)(dy/dt)/dx= (-1/x²)dy/dt+ (1/x)d(dy/dt)/dx= ((-1/x²)dy/dt+(1/x)(1/x)d²y/dt²=.
Now, x² y'' = d²y/dt²- dy/dt
and -3xy'= -3 dy/dt
so the differential equation is
d²y/dt²- 4dy/dt+ 13y= 4+ 3e^tt

That should be easy to solve!