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