I solved the first IVP, $dx/dt = -x$. The solution is $x(t)=pe^{-t}$. Now how should I solve $dy/dt=2y+x^2$. I tried by putting this value of $x(t)$ in $dy/dt$ and tried solving the ODE by finding an integrating factor. Here's what I did: $$y'(t)=2y(t)+p^2e^{-2t}$$ Using integrating factor, $y(t)=-(p^2/4)e^{-2t}+c/e^{-2t}$. Is this correct?
No, the solution of the first equation does not have an exponential factor $-2$. It should be $x(t)=pe^{-t}$.
You have to insert the square of $x$, $x(t)^2=p^2e^{-2t}$. Perhaps you did this, but you did not communicate this clearly.
It is possible that the resulting equation for the constant in $y(t)$ reads as $q=y(0)=-\frac14p^2+c$. You should document more steps to let us see if you did it right.
You can use the variation of parameters method: you know the solutions $y_h$ to the homogeneous equation $y_h'-2y_h=0$, then look for a particular solution $y_p = A(t)y_h(t)$ such that $y_p' = 2y_p+pe^{-2t}$.
Solve this for $A(t)$ and add it to $y_h$ and you will have your solution.
