Undetermined Coefficients non-homogenous part

Question

$$y''+ 3y'-4y = xe^{-x}+e^{-4x}$$

$$y_p(x)= Axe^{-x} + Be^{-4x}$$

This leads me to a false answer. Where am I doing wrong?

Answer 1

For the homogeneous part $$y''+ 3y'-4y = xe^{-x}+e^{-4x}$$ $$\implies r^2+3r-4=0 \implies (r-1)(r+4)=0 \implies r=1,-4$$ Therefore, $$y=c_1e^x+c_2e^{-4x}$$ for the inhomogenous part try the particular solution $$y_p=cxe^{-4x}+(ax+b)e^{-x}$$

Edit

I ended with $$-8ce^{-4x}+3ce^{-4x}=e^{-4x}$$ $$\implies c=-1/5$$

Your solution for $(A,B)$ is correct The final solution is

$$\boxed{ y(x)=c_1e^x+c_2e^{-4x}-\frac 15xe^{-4x}-\frac 16xe^{-x}-\frac 1{36}e^{-x}}$$