Finding the General Solution of an ODE: Help Needed!

[Benny]

Hi, can someone please help me with the following ODE? I need to find the general solution.

[tex]
y = xy' + \frac{1}{{y'}}
[/tex]

Rearranging I get a quadratic in dy/dx.

[tex]
x\left( {\frac{{dy}}{{dx}}} \right)^2 - y\left( {\frac{{dy}}{{dx}}} \right) + 1 = 0
[/tex]

[tex]
\frac{{dy}}{{dx}} = \frac{{y \pm \sqrt {y^2 - 4x} }}{{2x}}
[/tex]

I don't know what to do from this point nor am I sure if I've started the right way. Any help would be good thanks.

 

[J77]

That's a first order equation of degree two - you have to look at the discriminant you've found to find where solutions exist...

edit: I see sid deleted his reply - lucky I didn't quote it

 

[siddharth]

This is a Clairaut Equation, and there is a nice way to find the general and singular solutions.

http://mathworld.wolfram.com/ClairautsDifferentialEquation.html"

 

[Benny]

Now that you mention the name of the DE, I remember doing a question on it last year. It's too bad that I've put that booklet containing the problem and the books which I did questions in away in storage. Anyway thanks for the help. It looks like I need to diff both sides wrtx.