general and singular solution

• Jan 18th 2014, 08:28 AM
general and singular solution
How to proceed to find the general and singular solution of the equation
3xy=2px2-2p2, p=dy/dx
• Jan 18th 2014, 05:55 PM
chiro
Re: general and singular solution

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 PM
Prove It
Re: general and singular solution
When you write \displaystyle \begin{align*} p^2 \end{align*}, are you meaning \displaystyle \begin{align*} \left( \frac{dy}{dx} \right) ^2 \end{align*} or \displaystyle \begin{align*} \frac{d^2y}{dx^2} \end{align*}?
• Jan 18th 2014, 09:19 PM
Re: general and singular solution
(dy/dx)^2
• Jan 19th 2014, 12:32 AM
Prove It
Re: general and singular solution
Quote:

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

I would probably try something like this:

\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 AM