The Solution of $x\frac{{dy}}{{dx}} + y = {e^x}$ is

Given that $x\frac{{dy}}{{dx}} + y = {e^x}$

$\Rightarrow$ $\frac{{dy}}{{dx}} + \frac{y}{x} = \frac{{{e^x}}}{x}$

which is a linear differential equation.

$\therefore {\rm{IF}} = {e^{\int {\frac{1}{x}} dx}} = {e^{(\log x)}} = x$

The general Solution is $y \cdot x = \int {\left( {\frac{{{d^x}}}{x} \cdot x} \right)} dx$

$\Rightarrow$ $y \cdot x = \int {{e^x}} dx$

$\Rightarrow$ $y \cdot x = {e^x} + k$

$\Rightarrow$ $y = \frac{{{e^x}}}{x} + \frac{k}{x}$