Respuesta :
Answer: [tex]F(t) = 4 e ^ {\frac{t^{3} }{3}}-1[/tex]
Step-by-step explanation:
[tex]F'(t) = t^{2} (1+ F(t))[/tex]
[tex]\frac{dF}{dt} = t^{2} (1+F)[/tex]
First, we separated the variables:
[tex]\frac{dF}{1+F} = t^{2} dt[/tex]
We integrate:
[tex]\int\limits^{F(t)}_{F(0)} {\frac{1}{1+F'} } \, dF' = \int\limits^t_0 {t'^{3}} \, dt[/tex]
We change the variable for make more easy the integral:
[tex]u = 1+F', du = dF'[/tex]
[tex]\int\limits^{}_{} {\frac{1}{u} } \, du = \int\limits^0_t {t'^{3}} \, dt[/tex]
[tex]ln(u)= \frac{t^{3}}{3}[/tex]
Now replace with [tex]u = 1+F' [/tex] and evaluate the limit:
[tex]ln(1+F(t)) - ln (1+F(0)) = \frac{t^{3}}{3}[/tex]
Using properties of natural logarithm
[tex]ln(1+F(t))= \frac{t^{3}}{3} + ln(4)[/tex]
Taking exponential of both sides
[tex]e^{ln(1+F(t))} = e ^ {\frac{t^{3} }{3} + ln4}[/tex]
[tex]e^{ln(1+F(t))} = e ^ {\frac{t^{3} }{3}}e^{ln4} =e ^ {\frac{t^{3} }{3}} * 4[/tex]
[tex]1+F(t) = 4 e ^ {\frac{t^{3} }{3}}[/tex]
[tex]F(t) = 4 e ^ {\frac{t^{3} }{3}}-1[/tex]
