How to Solve a differential equation with both $x$ and $y$?

Question

Solve $\dfrac{dy}{dx}=\dfrac{y-3}{y^2+x^2}$ given that it passes through $(0,1)$.

Right now I do not yet know how to solve differential equations with both $x$ and $y$ that you cannot separate. Please help. Thanks.

Answer 1

Hint:

Let $y=u+3$ ,

Then $\dfrac{dy}{dx}=\dfrac{du}{dx}$

$\therefore\dfrac{du}{dx}=\dfrac{u}{(u+3)^2+x^2}$

$\dfrac{dx}{du}=\dfrac{x^2}{u}+\dfrac{(u+3)^2}{u}$

Let $x=-\dfrac{u}{v}\dfrac{dv}{du}$ ,

Then $\dfrac{dx}{du}=-\dfrac{u}{v}\dfrac{d^2v}{du^2}-\dfrac{1}{v}\dfrac{dv}{du}+\dfrac{u}{v^2}\left(\dfrac{dv}{du}\right)^2$

$\therefore-\dfrac{u}{v}\dfrac{d^2v}{du^2}-\dfrac{1}{v}\dfrac{dv}{du}+\dfrac{u}{v^2}\left(\dfrac{dv}{du}\right)^2=\dfrac{u}{v^2}\left(\dfrac{dv}{du}\right)^2+\dfrac{(u+3)^2}{u}$

$\dfrac{u}{v}\dfrac{d^2v}{du^2}+\dfrac{1}{v}\dfrac{dv}{du}+\dfrac{(u+3)^2}{u}=0$

$u^2\dfrac{d^2v}{du^2}+u\dfrac{dv}{du}+(u+3)^2v=0$

Hint: http://www.wolframalpha.com/input/?i=x%5E2y%22%2Bxy%27%2B%28x%2B3%29%5E2y%3D0

Answer 2

$$\dfrac{dy}{dx}=\dfrac{y-3}{y^2+x^2}$$ The equation is solvable but very arduous. If this is an academic exercise there is probably a typo. Where does the equation come from ?

I guess that editing all the calculus would be a waste of time. Probably the explicit solution which involves special functions such as confluent hypergeometric would be of no use. So, I give only a short view on a method to solve it.

Consider the inverse function $x(y)$. The equation becomes : $$\dfrac{dx}{dy}=\dfrac{1}{y-3}\big(x(y)\big)^2+\dfrac{y^2}{y-3}$$ This is a Ricatti equation (4) in https://mathworld.wolfram.com/RiccatiDifferentialEquation.html which is theoretically solvable thanks to the method shown.