Answer
$y=\ln \left(\frac{e}{1-ex^3}\right)$
Work Step by Step
Solve for $y$:
$\frac{dy}{dx}=3x^2e^{y}$ (Separate the variables)
$\frac{1}{e^{y}}dy=3x^2dx$
$-e^{-y}dy=-3x^2dx$ (Integrate)
$\int -e^{-y}dy=\int -3x^2 dx$
$e^{-y}=-x^3+C$
Substitute the initial value $y(0)=1$:
$e^{-1}=-0^3+C$
$e^{-1}=0+C$
$C=e^{-1}$
Now we have:
$e^{-y}=-x^3+e^{-1}$
$\frac{1}{e^y}=\frac{1-ex^3}{e}$
$e^{y}=\frac{e}{1-ex^3}$
Taking the logarithm,
$y=\ln (\frac{e}{1-ex^3})$
Here is the graph of this solution.