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I tried solving a system of two second order nonlinear ordinary differential equations using the DSolve command. First, I tried like this:

eqns = {A''[x] == 2/B[x]*A'[x]*B'[x], 
   B''[x] + 1/B[x]*(A'[x])^2 - 1/B[x]*(B'[x])^2 == 0};
sol = DSolve[eqns, {A, B}, x]

However, as Mathematica didn't (couldn't?) solve this, I transformed it into a system of four first order equations:

eqns = {c'[x] == 2/B[x]*c[x]*d[x], 
   d'[x] + 1/B[x]*(c[x])^2 - 1/B[x]*(d[x])^2 == 0, c[x] == A'[x], 
   d[x] == B'[x], c[0] == 1, d[0] == 1, A[0] == 1, B[0] == 1};
sol = DSolve[eqns, {A, B, c, d}, x]

This still doesn't work. Weirdly enough, I don't even get an error message.

I only included the boundary conditions thinking that they may be helpful, but they aren't part of my original problem.

Your help would be greatly appreciated:)

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    $\begingroup$ This is a non linear ODE, are you sure that a closed form solution exists? If I choose initial conditions and use "NDSolve" I can easily get numerical solutions. E.g. (A[0] == 1, B[0] == 1, A'[0] == .1, B'[0] == .1) $\endgroup$
    – Daniel Huber
    Apr 21, 2021 at 16:28
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    – bbgodfrey
    Apr 22, 2021 at 18:20

1 Answer 1

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With a bit of assistance, DSolve can produce a symbolic answer, as desired. Solve the first ODE for A'[x],

Equal@@(D[DSolve[eqns // First, A[x], x], x][[1, 1]])
(* A'[x] == B[x]^2 C[1] *)

and use this result instead of the first ODE

FullSimplify@DSolve[{%, eqns // Last}, {A[x], B[x]}, x] /. C[1] -> C[1] Sqrt[C[2]]
    /. Sqrt[Sech[z_]^2] -> Sech[z]
(* {{B[x] -> -Sech[(x - C[2]) Sqrt[C[2]]]/C[1], 
     A[x] -> C[4] + Tanh[(x - C[2]) Sqrt[C[2]]]/C[1]}, 
    {B[x] ->  [Sech[(x - C[2]) Sqrt[C[2]]]/C[1], 
     A[x] -> C[4] + Tanh[(x - C[2]) Sqrt[C[2]]]/C[1]}, 
    {B[x] -> -Sech[Sqrt[C[2]] (x + C[2])]/C[1], 
     A[x] -> C[4] + Tanh[Sqrt[C[2]] (x + C[2])]/C[1]}, 
    {B[x] ->  Sech[Sqrt[C[2]] (x + C[2])]/C[1], 
     A[x] -> C[4] + Tanh[Sqrt[C[2]] (x + C[2])]/C[1]}} *)

These solutions can be verified by

% /. Rule[z1_, z2_] :> Rule[Head[z1], Function[x, z2]];
FullSimplify[eqns /. %]
(* {{True, True}, {True, True}, {True, True}, {True, True}} *)
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