What is the solution for (tdu)/(dt)=t^2+3u,t>0,\H(5)=100 ?

The solution for (tdu)/(dt)=t^2+3u,t>0,\H(5)=100 is u=-t^2+c_{1}t^3

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(tdu)/(dt)=t^2+3u,t>0,\H(5)=100

\frac{t d u}{d t} = t^{2} + 3 u, t > 0, H (5) = 100

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{d}{d x} (x - 2 e^{x})

x y^{^{'}} + y^{^{'}} - (x + 1) sin (x) + y = 0

\frac{6}{s} + \frac{8 ( 4 )}{s ^{2} + 16}

e^{11}

f (x) = x - 6 3 x

What is the solution for (tdu)/(dt)=t^2+3u,t>0,\H(5)=100 ?
The solution for (tdu)/(dt)=t^2+3u,t>0,\H(5)=100 is u=-t^2+c_{1}t^3