Question

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

f(0) = 3

f(t) = ?

Answer 1

Answer: $F(t) = 4 e ^ {\frac{t^{3} }{3}}-1$

Step-by-step explanation:

$F'(t) = t^{2} (1+ F(t))$

$\frac{dF}{dt} = t^{2} (1+F)$

First, we separated the variables:

$\frac{dF}{1+F} = t^{2} dt$

We integrate:

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

We change the variable for make more easy the integral:

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

$\int\limits^{}_{} {\frac{1}{u} } \, du = \int\limits^0_t {t'^{3}} \, dt$

$ln(u)= \frac{t^{3}}{3}$

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

$ln(1+F(t)) - ln (1+F(0)) = \frac{t^{3}}{3}$

Using properties of natural logarithm

$ln(1+F(t))= \frac{t^{3}}{3} + ln(4)$

Taking exponential of both sides

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

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

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

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