F(prime)(t) = t^2 (1+f(t)) f(0) = 3 f(t) = ?

Question

asked Jul 26, 2020 58.9k views

1 Answer

Jeffkmeng · Answer 1 · 2020-08-01T11:55:50+0000

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

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

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

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

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