2
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Eq1:

$ \frac{1}{3\phi}=\left[\frac{x}{x+1}\frac{d^{2}}{dx^{2}}+\frac{2x+1}{(x+1)^{2}}\frac{d}{dx}\right]{\phi^{2}} $

Eq2:

$ \frac{b}{(x+1)^{2}}+\frac{{a}^{\prime}}{x+1}+\frac{b-a}{x(x+1)^{2}}=-\frac{1}{3\phi}+\left[\frac{x}{x+1}\frac{d^{2}}{dx^{2}}+\frac{1}{2(x+1)^{2}}\frac{d}{dx}\right]{\phi^{2}} $

Eq3:

$ \frac{b}{(x+1)^{2}}+\frac{{b}^{\prime}}{x+1}+\frac{1}{3\phi}=\frac{1}{2}\frac{1}{(x+1)^{2}}\frac{d}{dx}{\phi^{2}} $

where $a$, $b$ and $\phi$ are functions of $x$.

$a[0]=b[0]=\phi[0]=0$

I tried to plot the first equation alone with condition above using NDSolve but some error appeared

This is what I have done so far:

NDSolve[{x/(x + 1) D[\[Phi][x]^2, {x, 2}] + (
     2 x + 1)/(x + 1)^2 D[\[Phi][x], {x, 2}] == 1/(
   3 \[Phi][x]), \[Phi][0] == 0}, \[Phi], {x, 0, 0.02}]

But got this error

NDSolve::ndnco: The number of constraints (1) (initial conditions) is not equal to the total differential order of the system (2). >>

I think I have to include all three equations in NDSolve, but I don't know how to do that

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closed as off-topic by user9660, MarcoB, RunnyKine, m_goldberg, Öskå May 3 '16 at 20:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question cannot be answered without additional information. Questions on problems in code must describe the specific problem and include valid code to reproduce it. Any data used for programming examples should be embedded in the question or code to generate the (fake) data must be included." – Community, MarcoB, RunnyKine, Öskå
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Artes, I have added demonstrated what I have done so far $\endgroup$ – M007 Apr 4 '15 at 14:27
  • 1
    $\begingroup$ The equation you entered into the NDSolve is not equivalent to your Eq1. Eq1 would be entered as eq1 = x/(x + 1) D[\[Phi][x]^2, {x, 2}] + (2 x + 1)/(x + 1)^2 D[\[Phi][x]^2, x] == 1/(3 \[Phi][x]) $\endgroup$ – Bob Hanlon Apr 4 '15 at 15:09
  • $\begingroup$ I'm voting to close this question as off-topic because it is too localized; i.e, it applies only to the local situation and needs of its poster and answers will not benefit others. $\endgroup$ – m_goldberg May 2 '16 at 1:41