Question

Solve the following differential equation:
$4\frac{dy}{dx}+8y={5e}^{-3x}$

Answer 1

$4\frac{dy}{dx}+8y=5e^{-3x}$

$\frac{dy}{dx}+2y=\frac{5}{4}{e}^{-3x}$

$m+2=0$

$m=-2$

$A.E\Rightarrow y\left(x\right)=C_1{e}^{-2x}$

$P.I=\frac{5e^{-3x}}{4(D+2)}\Rightarrow \frac{5e^{-3x}}{4(-3+2)}=\frac{-5e^{-3x}}{4}$

C.S $y\left(x\right)=C_1{e}^{-2x}-\frac{5}{4}{e}^{-3x}$