- #1
Benny
- 584
- 0
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.
[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.