This ODE is linear.
\(t\dfrac{\mathrm du}{\mathrm du}-3u=t^2\)
Multiplying both sides by \(t^2\) gives
\(t^3\dfrac{\mathrm du}{\mathrm du}-3t^2u=t^4\)
\(\dfrac{\mathrm d}{\mathrm dt}[t^3u]=t^4\)
Integrating both sides with respect to \(t\) gives
\(t^3u=\displaystyle\int t^4\,\mathrm dt\)
\(t^3u=\dfrac15t^5+C\)
\(u=\dfrac15t^2+Ct^{-3}\)
Given that \(u(2)=4\), you have
\(4=\dfrac45+\dfrac C8\)
\(\implies C=\dfrac{128}5\)
