Question

solve the initial value problem y^(2)y'=xe^(-y^(3)) where y(0)=1

Answer 1

\[\large\begin{align*}
y^2y'&=xe^{-y^3}\\\\
y^2e^{y^3}~dy&=x~dx\end{align*}\]
Integrate both sides.

Answer 2

i get (1/3)e^(y^3)=(1/2)x^(2)+C but i can't figure out how to solve for y

Answer 3

Multiply both sides by 3, then take the logarithm. You can leave your answer in terms of \(y^3\), or take the cube root from there.