# differential eq of first order and higher degree

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

#### Ackbach

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

##### Member
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

##### Indicium Physicus
Staff member
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

##### Indicium Physicus
Staff member
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.

#### Jester

##### Well-known member
MHB Math Helper
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).

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