Question

Solve the initial value problem:
y' = x^3(1-y)
y(0)=3

Answer 1

\[\int\limits_{}^{} \frac{1}{1-y} dy= \int\limits_{}^{}x^{3} dx\]

Answer 2

solve by seperation of variables now use a u=1-y substitution and you should be able to finish it

Answer 3

\[u=1-y--------\int\limits \frac{1}{u} du= \ln(u) =\int\limits x^{3}=-\frac{x^{4}}{4}+C\]

Answer 4

\[1-y = e^{- \frac{x^{4}}{4}+c}\]

Answer 5

gj :)

Answer 6

Thanks