How to solve $\frac{dy}{dx}-1=\frac{x^2}{y}$ How to solve $$\frac{dy}{dx}-1=\frac{x^2}{y}$$
I tried letting $y^2=u$ which gives $$y\frac{dy}{dx}=\frac{1}{2}\frac{du}{dx}$$
So the equation is now
$$\frac{du}{dx}-2\sqrt{u}=2x^2$$
Any help?
 A: Not an answer
As a possible  aid to others in possibly guessing analytical solution curve shapes, integrands of
$$\frac{dy}{dx}-c=\frac{x^2}{y}$$
with initial slope values $ y^{'}_{i}=c $ at $x=0 $ are numerically computed and plotted. When $c=0,$ $y\to k\cdot x^{3/2}$
( fwiw with no possible further consequence its second order ode .. if I made no error in differentiating is)
$$ \frac{y y'''-y''}{y'y''}= (2 c-1). $$

A: After reviewing comments:
use:
$x=\cos{\theta},\;y=\sin{\theta}$
Then:
$dy=\cos{\theta}d\theta,\;dx=-\sin{\theta}d\theta$.
Substitute in :
$\frac{dy}{dx}=1+\frac{x^2}{y}$
$\cos{\theta}d\theta=-\sin{\theta}d\theta+\frac{\cos^2{\theta}}{\sin{\theta}}\cdot(-\sin{\theta})d\theta$
So now everything is integrable.
$(\cos{\theta}+\sin{\theta}+\frac{1}{2}(1+\cos{2\theta}))d\theta=0$
$-\sin{\theta}+\cos{\theta}+\frac{1}{2}\theta+\frac{1}{4}\sin{2\theta}=c$
Substitute back.
$-y+x+\frac{1}{2}\arctan{\frac{y}{x}}+\frac{1}{2}xy=c$.
$c$ constant to be determined from initial condition.
