What is the solution for (tdu)/(dt)=t^2+3u,t>0,\H(6)=180 ?

The solution for (tdu)/(dt)=t^2+3u,t>0,\H(6)=180 is u=-t^2+c_{1}t^3

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(tdu)/(dt)=t^2+3u,t>0,\H(6)=180

\frac{t d u}{d t} = t^{2} + 3 u, t > 0, H (6) = 180

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}

5 - 2 x^{2}

\frac{d y}{d x} = 0.05 x - 450

\frac{0.6}{s + 10}

\int \frac{e ^{x} - 3}{x} d x

x \to 1 lim (2 x^{2} - 3 x + 4)

What is the solution for (tdu)/(dt)=t^2+3u,t>0,\H(6)=180 ?
The solution for (tdu)/(dt)=t^2+3u,t>0,\H(6)=180 is u=-t^2+c_{1}t^3