Answer
$$ y=\frac{1}{3}x^3+1.$$
$$ y= -\frac{1}{x-1}.$$
Work Step by Step
For the equation
$$\frac{dy}{dx}=x^2\Longrightarrow y=\frac{1}{3}x^3+c.$$
Using the condition $ y(0)=1$, we obtain that $1=0+c $; then $ c=1$ and hence the solution is $$ y=\frac{1}{3}x^3+1.$$
For the equation
$$\frac{dy}{dx}=y^2\Longrightarrow \frac{dy}{y
^2}=dx \Longrightarrow - \frac{1}{y} =x+c.$$
Using the condition $ y(0)=1$, we obtain that $-1=0+c $; then $ c=-1 $ and hence the solution is $$ y= -\frac{1}{x-1}.$$
