Question

F(prime)(t) = t^2 (1+f(t))

f(0) = 3

f(t) = ?

Answer 1

Answer:
`F(t) = 4 e ^ {(t^(3) )/(3)}-1`

Explanation:

`F'(t) = t^(2) (1+ F(t))`

`(dF)/(dt) = t^(2) (1+F)`

First, we separated the variables:

`(dF)/(1+F) = t^(2) dt`

We integrate:

`\int\limits^(F(t))_(F(0)) {(1)/(1+F') } \, dF' = \int\limits^t_0 {t'^(3)} \, dt`

We change the variable for make more easy the integral:

`u = 1+F', du = dF'`

`\int\limits^{}_{} {(1)/(u) } \, du = \int\limits^0_t {t'^(3)} \, dt`

`ln(u)= (t^(3))/(3)`

Now replace with
`u = 1+F'` and evaluate the limit:

`ln(1+F(t)) - ln (1+F(0)) = (t^(3))/(3)`

Using properties of natural logarithm

`ln(1+F(t))= (t^(3))/(3) + ln(4)`

Taking exponential of both sides

`e^(ln(1+F(t))) = e ^ {(t^(3) )/(3) + ln4}`

`e^(ln(1+F(t))) = e ^ {(t^(3) )/(3)}e^(ln4) =e ^ {(t^(3) )/(3)} * 4`

`1+F(t) = 4 e ^ {(t^(3) )/(3)}`

`F(t) = 4 e ^ {(t^(3) )/(3)}-1`