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differential eq of first order and higher degree

suvadip

Member
Feb 21, 2013
69
How to proceed to find the general and singular solution of the equation
3xy=2px2-2p2, p=dy/dx
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Question: does $p^{2}$ mean $\displaystyle \left( \frac{dy}{dx} \right)^{ \! 2}$ or $\displaystyle \frac{d^{2}y}{dx^{2}}$?
 

suvadip

Member
Feb 21, 2013
69
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
Jan 26, 2012
4,197
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
Jan 26, 2012
4,197
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
Jan 26, 2012
183
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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