Answer
$x=-t^2+C_1t^3+C_2$
Work Step by Step
We are given:
$t\frac{d^2x}{dt^2}=2(t+\frac{dx}{dt})$
Suppose that
$u=\frac{dx}{dt}$
Now, $\frac{du}{dt}=\frac{d^2x}{dt^2}$
Subsitute to the equation:
$t\frac{du}{dt}=2(t+u)$
$\frac{du}{dt}-\frac{2u}{t}=2$
Integrating factor:
$I(x)=e^{\int -\frac{2}{t}dx}=e^{-2\ln t}=t^{-2}$
The equation becomes:
$\frac{d}{dx}(ut^{-2})=2t^{-2}$
Integrating both sides:
$ut^{-2}=-2t^{-1}+C$
$ u=-2t+Ct^2$
$\frac{dx}{dt}=-2t+Ct^2$
$dx=-2t+Ct^2dt$
Integrating the last equation,
$x=-t^2+C_1t^3+C_2$
The final solution is:
$x=-t^2+C_1t^3+C_2$
