Question

Find the solution of the ordinary differential equation t(du/dt)=t^2+3u, t greater than 0. u(2)=4

Answer 1

This ODE is linear.


`t(\mathrm du)/(\mathrm du)-3u=t^2`

Multiplying both sides by
`t^2` gives


`t^3(\mathrm du)/(\mathrm du)-3t^2u=t^4`

`(\mathrm d)/(\mathrm dt)[t^3u]=t^4`

Integrating both sides with respect to
`t` gives


`t^3u=\displaystyle\int t^4\,\mathrm dt`

`t^3u=\frac15t^5+C`

`u=\frac15t^2+Ct^(-3)`

Given that
`u(2)=4`, you have


`4=\frac45+\frac C8`

`\implies C=\frac{128}5`