Solve the differential equation to get $y $ as a function of $x$.

$$x\frac{dy}{dx} + 3\frac{dx}{dy} = y^2$$

I tried taking $\dfrac{dy}{dx}$ to be $p$ . Then proceed to factorise the expression but nothing happened. Please help me in this question as I don't know how to proceed.


We change the role of $x$ and $y$, therfore $$y\dfrac{dx}{dy}+3\dfrac{dy}{dx}=x^2$$ gives the equation $$x^2y'-3y'^2=y$$ which has two following roots $$y'=\dfrac{-x^2\pm\sqrt{x^4-12y}}{-6}$$ now let $z^2=x^4-12y$ which leads us to DE $$(2x^3-x^2+z)dx-zdz=0$$ that could be convert to an exact equation with $\dfrac{\partial M}{\partial z}=1$ and $\dfrac{\partial N}{\partial x}=0$ (with standard notation). After solve that and substitution, exchange the previous role of $x$ and $y$.

  • $\begingroup$ Thank you very much sir for the method. Nice approach. $\endgroup$ – Be happy Jul 29 '18 at 3:03
  • $\begingroup$ Sorry. I couldn't solve it completely! $\endgroup$ – Nosrati Jul 29 '18 at 12:30

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