Answer
\[x(t)=\frac{C+4e^t(t-1)}{t^2}\]
Work Step by Step
$t\large\frac{dx}{dt}+\small 2x=4e^t$ _____(1)
$\large\frac{dx}{dt}+\large\frac{2}{t}\small x=\large\frac{4e^t}{t}$ ___(2)
(1) is linear differential equation,
Integrating factor:-
$I(t)=e^{\int\large\frac{2}{t}}dt=e^{2\ln t}=t^2$
Multiply (2) by $I(t)$
\[t^2\frac{dx}{dt}+2xt=4e^tt\]
\[\frac{d}{dt}[xt^2]=4te^t\]
Integrating,
\[xt^2=4\int te^tdt+C\]
$C$ is constant of integration
\[xt^2=4e^t(t-1)+C\]
\[x(t)=\frac{C+4e^t(t-1)}{t^2}\]
Hence general solution of (1) is $x(t)=\frac{C+4e^t(t-1)}{t^2}$
