# Math Help - general and singular solution

1. ## general and singular solution

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

2. ## 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)

3. ## 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*}?

(dy/dx)^2

5. ## Re: general and singular solution

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...

6. ## Re: general and singular solution

how to find y from the last expression?

7. ## Re: general and singular solution

That is NOT a 'linear' differential equation and so may NOT have a "general" solution at all.

8. ## Re: general and singular solution

y=px+p^2 is also not a linear diff eq, but its general solution is y=cx+c^2