Answer
$y=\dfrac{x^4}{4}-x^2+1$
Work Step by Step
We are given that $\dfrac{dy}{dx}=x^3-2x$
We will separate the variables to obtain:
$ \ dy=(x^3-2x) \ dx$
Integrate to obtain:
$\int \ dy=\int (x^3-2x) \ dx$
This implies that $y=\dfrac{x^4}{4}-x^2+C$
After applying the initial conditions. $y=1$ when $x=0$, we get $C=1$
Therefore, we have: $y=\dfrac{x^4}{4}-x^2+1$
