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I have this differential equation, for a function $ h(x) $ subject to a initial condition $ h(0) = n $, and a relation $ h^2(x) = p(x) $

$$ (1 + x)^4 \frac{\mathrm{d} }{\mathrm{d} x} \left(\frac{h(x)^2}{(1 + x)^3}\right) = 2 h(x) $$

I solve it with Mathematica 

DSolve[{(1 + x)^4 D[h[x]^2/(1 + x)^3, x] == 2 h[x], h[0] == n}, h, {x,0, 1150}]

And I get this solution 

h -> Function[{x}, 1/3 (-2 + 2 (1 + x)^(3/2) + 3 n (1 + x)^(3/2))

If I use the relation for $ h(x) $, I get this differential equation

$$ (1 + x)^4 \frac{\mathrm{d}}{\mathrm{d} x} \left(\frac{p(x)}{(1 + x)^3}\right) = 2 \sqrt{p(x)} $$

Solving with Mathematica, with the initial condition $ p(0) = n^2 $

DSolve[{(1 + x)^4 D[p[x]/(1 + x)^3, x] == 2 p[x]^(1/2), p[0] == n^{2}}, p, {x, 0, 1150}]

I get a different solution (Mathematica gives four solutions)

p -> Function[{x}, 1/9 (8 + 12 n + 9 n^2 + 12 x + 36 n x + 27 n^2 x + 12 x^2 + 36 n x^2 + 27 n^2 x^2 + 4 x^3 + 12 n x^3 + 9 n^2 x^3 - 8 Sqrt[1 + x] - 12 n Sqrt[1 + x] - 8 x Sqrt[1 + x] -12 n x Sqrt[1 + x])

Why we get two different solutions?

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    $\begingroup$ They seem to be equivalent. h[x]^2 - p[x] /. {hsol, psol} yields 0. $\endgroup$
    – Michael E2
    Commented Apr 14, 2020 at 3:57
  • $\begingroup$ Thank you for your answer. You know why mathematica gives four different solutions? $\endgroup$
    – No name
    Commented Apr 14, 2020 at 4:01
  • $\begingroup$ since you are using DSolve, why do you gives {x,0, 1150} instead of just x? $\endgroup$
    – Nasser
    Commented Apr 14, 2020 at 4:12
  • $\begingroup$ Oh, I was making some tests with NDSolve and I forget to delete the interval {x,0,1150} $\endgroup$
    – No name
    Commented Apr 14, 2020 at 4:16
  • $\begingroup$ No, not off hand, other than it seems to get a solution implicitly as an equation, from which Solve [sic] returns four branches. Only one of them has only real coefficients. $\endgroup$
    – Michael E2
    Commented Apr 14, 2020 at 4:19

1 Answer 1

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Solve for initial condition later and omit solution = zero.

dsol = DSolve[{(1 + x)^4 D[h[x]^2/(1 + x)^3, x] == 2 h[x]}, h, x]

(*   {{h -> Function[{x}, 0]}, 
      {h -> Function[{x}, -(2/3) + (1 + x)^(3/2) C[1]]}}   *)

sol = Solve[n == h[0] /. dsol[[2]], C[1]]

(*   {{C[1] -> 2/3 + n}}   *)

hh[x_] = h[x] /. dsol[[2]] /. sol[[1]]

(*   -(2/3) + (2/3 + n) (1 + x)^(3/2)   *)

p[x_] = hh[x]^2 // Simplify

(*   (-(2/3) + (2/3 + n) (1 + x)^(3/2))^2   *)

p[0]

(*   n^2   *)
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  • $\begingroup$ Thank you for your recomendation $\endgroup$
    – No name
    Commented Apr 14, 2020 at 5:15

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