MHB Differential eq of first order and higher degree

[Suvadip]

How to proceed to find the general and singular solution of the equation
3xy=2px²-2p², p=dy/dx

 

[Ackbach]

Question: does $p^{2}$ mean $\displaystyle \left( \frac{dy}{dx} \right)^{ \! 2}$ or $\displaystyle \frac{d^{2}y}{dx^{2}}$?

 

[Suvadip]

Question: does $p^{2}$ mean $\displaystyle \left( \frac{dy}{dx} \right)^{ \! 2}$ or $\displaystyle \frac{d^{2}y}{dx^{2}}$?

P^2=(dy/dx)^2

 

[Ackbach]

P^2=(dy/dx)^2

So, in that case, one thing you can try is simply solve for the derivative algebraically first:
\begin{align*}
3xy&=2x^{2} \frac{dy}{dx}-2 \left( \frac{dy}{dx} \right)^{ \! 2} \\
0&=2 \left( \frac{dy}{dx} \right)^{ \! 2}-2x^{2} \frac{dy}{dx}+3xy \\
\frac{dy}{dx} &= \frac{2x^{2} \pm \sqrt{4x^{4}-4(8)(3xy)}}{4} \\
&= \frac{x^{2} \pm \sqrt{x^{4}-24xy}}{2}.
\end{align*}
Unfortunately, either of the resulting DE's,
$$ \frac{dy}{dx}=\frac{x^{2} + \sqrt{x^{4}-24xy}}{2}$$
or
$$ \frac{dy}{dx}=\frac{x^{2} - \sqrt{x^{4}-24xy}}{2}$$
seem rather forbidding. They might be homogeneous, though. Try that.

[EDIT] Never mind about the homogeneous bit. Neither resulting DE is homogeneous.

 

[Ackbach]

Mathematica yields $$\{ \{ {{y(x)}\rightarrow
{\frac{-\left( e^{\frac{3\,C(1)}{2}}\,
\left( 3\,e^{\frac{3\,C(1)}{2}} -
{\sqrt{6}}\,x^{\frac{3}{2}} \right) \right) }{3}}}\} ,
\{ {{y(x)}\rightarrow
{\frac{-\left( e^{\frac{3\,C(1)}{2}}\,
\left( 3\,e^{\frac{3\,C(1)}{2}} +
{\sqrt{6}}\,x^{\frac{3}{2}} \right) \right) }{3}}}\} \}.$$

Also, by inspection, you can see that $y=0$ solves the DE. From the looks of the solutions above, this might be a singular solution.

 

[Mathwebster]

Divide your equation by $x$ and then differentiate. You should find that the new equation factors into two pieces that integrate easily. With these, go back to the original equation and check that they both work (and adjust constants accordingly).