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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

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

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

Mathematica graphics

Lets check what happens to s at the above boundary condition

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

Mathematica graphics

The same for the other boundary condition

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

Mathematica graphics

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

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

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