Question

Solve: $\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+e^x=0$

Answer 1

$\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+e^x=0\Rightarrow \frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=-e^x$

Auxillary equation $m^2+m=0$

$m(m+1)=0$

$m=0,m=-1$

$y_h\left(x\right)=C_1{e}^{0x}+C_2{e}^{-x}\Rightarrow y_h\left(x\right)=C_1+C_2{e}^{-x}$

$y_p\left(x\right)=A{e}^x$

$y_{p^{\prime }}\left(x\right)=A{e}^x$

Putting the value of $y_p$ and$Y_{p^{\prime }}$ in the differential equation we get

$A{e}^x+A{e}^x=-e^x$

$2A{e}^x=-e^x\Rightarrow 2A=-1\Rightarrow A=-\frac{1}{2}$

$\therefore y_p\left(x\right)=-\frac{1}{2}{e}^x$

$\therefore y\left(x\right)=y_h\left(x\right)+y_p\left(x\right)$

$y\left(x\right)=C_1+C_2{e}^{-x}+(-\frac{1}{2})e^x$