Question

Solve the initial-value problem. Differential equations.

Answer 1

\[t\frac{du }{ dt }=t^2+3u, t > 0, u(6) = 180\]

Answer 2

\[\frac{ du }{ dt }=t+\frac{ 3u }{ t }=> \frac{ du }{ dt }+(-\frac{ 3 }{ t }u)=t\]\[I(t) = e^{\int\limits_{}^{}(\frac{ -3 }{ t })dt}=e^{-3lnt}=e^{lnt^{-3}}=t^{-3}=\frac{ 1 }{ t^3 }\]\[\frac{ 1 }{ t^3 }*\frac{ du }{ dt }-\frac{ 3 }{ t^4 }=\frac{ 1 }{ t^2 }\]

What shall I do now?

Answer 3

Solve the differential equation, just like the others you've been doing.

Then, when you have u(t), plug in t=6 and u =180 and solve for the integration constant.

Answer 4

Ah okay, got it, thanks! :D