What is the general solution of
y'' + 4y' + 3y = 4e^(-2t)?

The answer is: C1e^(-t) + C2e^(-3t) - 4e^(-2t), I just don't know why the last part is subtracted and not added.

Question

What is the general solution of
 y'' + 4y' + 3y = 4e^(-2t)?

The answer is: C1e^(-t) + C2e^(-3t) - 4e^(-2t), I just don't know why the last part is subtracted and not added.

nikvist · Answer

$$y''+4y'+3y=4e^{-2t}\quad,\quad y=y_H+y_P$$$$y_H=C_1e^{\lambda_1 t}+C_2e^{\lambda_2 t}\quad,\quad\lambda^2+4\lambda+3=(\lambda+1)(\lambda+3)=0$$$$y_H=C_1e^{-t}+C_2e^{-3t}$$$$y_P=Ae^{-2t}\quad,\quad y'_P=-2Ae^{-2t}\quad,\quad y''_P=4Ae^{-2t}$$$$y''_P+4y'_P+3y_P=4Ae^{-2t}+4\cdot(-2Ae^{-2t})+3Ae^{-2t}=-Ae^{-2t}$$$$\Rightarrow\quad A=-4\quad\Rightarrow\quad y_P=-4e^{-2t}$$$$y=y_H+y_P=C_1e^{-t}+C_2e^{-3t}-4e^{-2t}$$

anonymous · Answer

ohhh great! i knew I was missing something! Thanks!