# How to solve this second-order linear differential equation?

$$y''(x)+4y'(x)+3y(x)=9x^2-20+30e^{2x}, \quad y(0)=0, \quad y'(0)=2$$

I'm stuck on finding a particular integral for the non-homogeneous R.H.S.: $9x^2 − 20 + 30e^{2x}$.

So $y= y_c + y_p$. I found $y_c = C_1e^{-x} + C_2e^{-3x}$ (where $C_1$ and $C_2$ are constants) But what would I try for $y_p$? If it helps, I know how to do it if it's just a polynomial (i.e. for a 2nd degree polynomial): you set $y_p= Ax^2 + Bx + C$ and $y'_p= Ax + B$ and $y''_p= A$

But in this case we have different things. What would I need to do? I would really appreciate your help. Thanks in advance!

Hint:

Let $$y_p=ax^2+bx+c+ke^{2x}$$ substitute this in the equation and proceed through.

Solving this

$$y"(x) + 4y'(x) + 3y(x) = 9x^2 − 20 + 30e^(2x) , y(0) = 0, y'(0) = 2$$

Is the same as solving $$y"(x) + 4y'(x) + 3y(x) = 9x^2 − 20$$

using $y_p=Ax^2+Bx+C$ and

$$y"(x) + 4y'(x) + 3y(x) = 30e^{2x}$$

using $y_p=Re^{2x}$

The solution is just going to be the sum of all particular solutions

$$y=y_h+y_{p1}+y_{p2}.....$$