Find the general solution of $\frac{dy}{dx} + ay = e^{mx}$.

$y = \frac{e^{mx}}{m + a} + Ke^{-ax}$

The correct answer is Option (2) → $y = \frac{e^{mx}}{m + a} + Ce^{-ax}$ ##

Given differential equation is

$\frac{dy}{dx} + ay = e^{mx}$

which is a linear differential equation.

On comparing it with $\frac{dy}{dx} + Py = Q$, we get

$P = a, Q = e^{mx}$

$\text{I.F.} = e^{\int P dx} = e^{\int a dx} = e^{ax}$

The general solution is $y \cdot e^{ax} = \int e^{mx} \cdot e^{ax} dx + C$

$\Rightarrow y \cdot e^{ax} = \int e^{(m + a)x} dx + C$

$\Rightarrow y \cdot e^{ax} = \frac{e^{(m + a)x}}{(m + a)} + C$

$\Rightarrow (m + a)y = \frac{e^{(m + a)x}}{e^{ax}} + \frac{(m + a)C}{e^{ax}}$

$\Rightarrow (m + a)y = e^{mx} + K e^{-ax} \quad [∵K = (m + a)C]$