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

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]