# How to solve this second order differential equation?

The equation is given by:

$$\frac{dv}{du}\left(2 u\frac{1-uv}{1+uv} + u\right) = \frac{v}{2} \frac{d^2v}{du^2} + v\frac{1-uv}{1+uv}$$

I tried using DSolve for this differential equation, but it's not giving me any answer.

DSolve[
v'[u] (2 u (1 - u v[u])/(1 + u v[u]) + u) ==
v[u] (1 - u v[u])/(1 + u v[u]) + v''[u] (v[u]/2),
v[u], u]


How do I solve this?

### Edit

In case you want to solve the equation numerically, the boundary conditions are

$$v\left(\frac{1+\sqrt{4.9}}{2}\right) = \frac{-2}{1 + \sqrt{4.9}}$$ $$v\left(\frac{1-\sqrt{4.9}}{2}\right) = \frac{-2}{1 - \sqrt{4.9}}$$

• is there any initial conditions !? – Alrubaie Apr 27 '19 at 15:37
• @Alrubaie There are, but the equation can be solved without the same, right? – Bruce Lee Apr 27 '19 at 15:58
• @Alrubaie I added the boundary conditions. – Bruce Lee Apr 27 '19 at 16:23
• Maybe ask at math.stackexchange.com and have the master integrators have a go? – Roman Apr 27 '19 at 18:37

In case you want to solve the equation numerically, the boundary conditions are

Your boundary conditions are not consistent with the ODE. Did you check?

Looking at this term in your ODE

 s = (1 - u*v[u])/(1 + u*v[u]);


Then you say one boundary condition is

 z = Sqrt[49/10];
(v[(1 + z)/2] == -2/(1 + z)) // N


Lets check what happens to s at the above boundary condition

 s /. {u -> (1 + z)/2, v[u] -> -2/(1 + z)}


The same for the other boundary condition

 s /. {u -> (1 - z)/2, v[u] -> -2/(1 - z)}


So you can not solve it even numerically unless you fix the boundary conditions.

• Thanks for the answer. :) It "seems" that the equation can be numerically solved with proper boundary conditions. If not, I will post the question. – Bruce Lee Apr 28 '19 at 9:52