How to solve $\frac{dy}{dx}-1=\frac{x^2}{y}$

Question

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?

Answer 1

Here the method of successive approximations can be used.

We denote the initial condition as

$$y_{0}=y(x_{0})$$

To obtain the zero approximation, we solve the equation: $$\frac{dy}{dx}-1=0$$ The solution:

$$y=x+c$$

where $$c=y_{0}-x_{0}$$

To obtain a first approximation, we substitute the obtained solution into the original differential equation: $$\frac{dy}{dx}-1=\frac{x^2}{x+c}$$

The solution: $$y=x(1-c)+\frac{x^2}{2}+c^2\ln (x+c)+c_{1}$$

To determine $c_{1}$ we use the initial condition again

$$c_{1}=(1+x_{0})c-x_{0}-\frac{x_{0}^2}{2}-c^2\ln y_{0}$$

To obtain a second approximation, we substitute the last solution into the original differential equation.

But unfortunately now it is not possible to obtain a simple analytical expression.

But for values ​​of $x$ not very far from $x_{0}$, this is a more or less satisfactory solution.

Answer 2

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)

$$ \frac{y y'''-y''}{y'y''}= (2 c-1). $$