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}

\frac{9}{x ^{4}}

\int tan (\frac{x}{2}) d x

\frac{\partial}{\partial x} (x^{2} + y cos (x y))

\int 4 sec^{4} (x) tan (x) d x

\frac{d}{d x} (\frac{x ^{5}}{5} + \frac{1}{12 x ^{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