What is the solution for (tdu)/(dt)=t^2+3u,\H(4)=48 ?

The solution for (tdu)/(dt)=t^2+3u,\H(4)=48 is u=-t^2+c_{1}t^3

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(tdu)/(dt)=t^2+3u,\H(4)=48

\frac{t d u}{d t} = t^{2} + 3 u, H (4) = 48

u = - t^{2} + c_{1} t^{3}

Solution steps

\frac{t d u}{d t} = t^{2} + 3 u

Solve linear ODE:

u = - t^{2} + c_{1} t^{3}

u = - t^{2} + c_{1} t^{3}

What is the solution for (tdu)/(dt)=t^2+3u,\H(4)=48 ?
The solution for (tdu)/(dt)=t^2+3u,\H(4)=48 is u=-t^2+c_{1}t^3