I want to sketch the phase portrait (solution curves) of this DE

$\displaystyle y'=y-x^2$

Could I use isoclines? i.e. something like this

$\displaystyle 0=y-x^2$

$\displaystyle =>y=x^2$

So i would sketch a parabola with slope 0 hairs on it? Then do the same thing for slope 1.

$\displaystyle 1=y-x^2$

$\displaystyle =>y=\frac{1}{x^2}$

So i would sketch 1/x^2 with slope 1 hairs on the curve?