How to proceed to find the general and singular solution of the equation

3xy=2px^{2}-2p^{2}, p=dy/dx

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- Jan 18th 2014, 08:28 AMSuvadipgeneral and singular solution
How to proceed to find the general and singular solution of the equation

3xy=2px^{2}-2p^{2}, p=dy/dx - Jan 18th 2014, 05:55 PMchiroRe: general and singular solution
Hey Suvadip.

For those unfamiliar with non-linear DE's, what techniques have you covered in your course? (Analytic as opposed to numeric which could be applied here) - Jan 18th 2014, 08:03 PMProve ItRe: general and singular solution
When you write $\displaystyle \displaystyle \begin{align*} p^2 \end{align*}$, are you meaning $\displaystyle \displaystyle \begin{align*} \left( \frac{dy}{dx} \right) ^2 \end{align*}$ or $\displaystyle \displaystyle \begin{align*} \frac{d^2y}{dx^2} \end{align*}$?

- Jan 18th 2014, 09:19 PMSuvadipRe: general and singular solution
(dy/dx)^2

- Jan 19th 2014, 12:32 AMProve ItRe: general and singular solution
I would probably try something like this:

$\displaystyle \displaystyle \begin{align*} 3x\,y &= 2x^2\,\frac{dy}{dx} - 2\left( \frac{dy}{dx} \right) ^2 \\ 2 \left( \frac{dy}{dx} \right) ^2 - 2x^2\,\frac{dy}{dx} + 3x\,y &= 0 \\ \left( \frac{dy}{dx} \right) ^2 - x^2\,\frac{dy}{dx} + \frac{3}{2}x\,y &= 0 \\ \left( \frac{dy}{dx} \right) ^2 - x^2 \, \frac{dy}{dx} + \left( -\frac{1}{2}x^2 \right) ^2 - \left( -\frac{1}{2}x^2 \right) ^2 + \frac{3}{2}x\,y &= 0 \\ \left( \frac{dy}{dx} - \frac{1}{2}x^2 \right) ^2 - \frac{1}{4}x^2 + \frac{3}{2}x\,y &= 0 \\ \left( \frac{dy}{dx} - \frac{1}{2}x^2 \right) ^2 &= \frac{x^2 - 6x\,y}{4} \\ \frac{dy}{dx} - \frac{1}{2}x^2 &= \frac{\pm \sqrt{x^2 - 6x\,y}}{2} \\ \frac{dy}{dx} &= \frac{x^2 \pm \sqrt{x^2 - 6x\,y}}{2} \end{align*}$

I don't know if this looks any easier to solve... - Jan 19th 2014, 02:20 AMSuvadipRe: general and singular solution
how to find y from the last expression?

- Jan 21st 2014, 05:19 AMHallsofIvyRe: general and singular solution
That is NOT a 'linear' differential equation and so may NOT have a "general" solution at all.

- Jan 23rd 2014, 08:25 PMSuvadipRe: general and singular solution
y=px+p^2 is also not a linear diff eq, but its general solution is y=cx+c^2