Hello Neil;

You are absolutely right, numerical solution would be a good way to go. But... The problem with numerical solution is that, i need to equate 2 nonlinear differential equations at a Point X to find X. Inorder to do this I have to have 2 analytical solutions at hand to equate. With numerical solution, I dont have any equation at hand. Unfortunately the analytic solutions of these differential equations involve inverse Elliptic Integrals. A polynomial approximation for the inverse elliptical integral (should be inverse) would work perfect for me. In your manual, it only gives direct approximation for the elliptic integral not the inverse approximation.

It would be easier and probably more reliable to get the numeric solution on some interval and try to convery the interpolating function to C code.

Maybe this helps a little:

Internal`InheritedBlock[{Solve}, Unprotect[Solve];
Solve[x___] := 
Block[{$guard = True}, Print["Solve called : ", HoldForm[Solve[x]]];
Solve[x]] /; ! TrueQ[$guard];
DSolve[{-x^2 + (0.232866 - 0.232166*Cos[y[x]])*y'[x]^2 == 0}, y[x], x]];

(*Solve called : Solve[10 Sqrt[14] EllipticE[y[x]/2,-(116083/175)]==-((5714200963 x^2)/16162201)+Subscript[\[ConstantC], 1],y[x]]*)

We see that solution is implicit form:

10 Sqrt[14] EllipticE[y[x]/2,-(116083/175)]==-((5714200963 x^2)/16162201)+C[1]

where: C[1] is a integration constant.

For: EllipticE,#and & see in Documentation Center

Yes you are correct. We can equation write:

HoldForm[Integrate[(1 + (116083/175)*Sin[t]^2)^(1/2), {t, 0, 
    y[x]/2}] == (-((5714200963 x^2)/16162201) + C[1])/(10 Sqrt[14])] //  TeXForm

$$\int_0^{\frac{y(x)}{2}} \sqrt{1+\frac{116083 \sin ^2(t)}{175}} \, dt=\frac{-\frac{5714200963 x^2}{16162201}+c_1}{10 \sqrt{14}}$$

Try this:

DSolve[{-x^2 + (0.232866 - 0.232166*Cos[y[x]])*y'[x]^2 == 0, y[2] == 1.57}, y[x], x]
(*{{y[x] -> 
   InverseFunction[10 Sqrt[14] EllipticE[#1/2, -(116083/175)] &][
    1699.28 - 353.553 x^2]}, {y[x] -> 
   InverseFunction[
     10 Sqrt[14] EllipticE[#1/2, -(116083/175)] &][-1129.14 + 
     353.553 x^2]}}*)

Then we extracting from outputs two solution:

  {10 Sqrt[14] EllipticE[y[x]/2, -(116083/175)] == 
    1699.2844439496437` - 353.5533905932738` x^2, 
   10 Sqrt[14]
      EllipticE[y[x]/2, -(116083/175)] == -1129.1426807965465` + 
     353.5533905932738` x^2}

No,It's mathematicaly impossible,because you have a transcendental equation to solve. With Mathematica is possible to plot solution from Dsolve:

 p1 = ContourPlot[{10 Sqrt[14] EllipticE[y/2, -(116083/175)] == 
     1699.2844439496437` - 353.5533905932738` x^2, 10 Sqrt[14]
      EllipticE[y/2, -(116083/175)] == -1129.1426807965465` + 
      353.5533905932738` x^2}, {x, -3, 3}, {y, -4, 4}, 
   FrameLabel -> Automatic]