How to get special solution with DSolve?

Question

I am trying to solve the following ODE with DSolve, but can't obtain a special solution (which I know exists simply by looking at the equation: f[x]=1)

a[x_] = f[x] /. 
  DSolve[{m f[x]^(3/2) f'[x] + 2 k f'[x]^2 - 2 k f[x] f''[x] == 0}, f[x], x]
Solve[a[0] == 1, C[1]]

The Solve parts simply gives me empty brackets.

If I include the boundary condition in DSolve I only get the error message "For some branches of the solution, unable to solve the conditions".

I know this is similar to other questions here, but I wasn't able to use the corresponding answers for my problem.

How I can resolve this?

Answer 1

The output of DSolve is a list

DSolve[{m f[x]^(3/2) f'[x] + 2 k f'[x]^2 - 2 k f[x] f''[x] == 0}, 
 f[x], x]

{{f[x] -> (
   E^(x C[1] + C[1] C[2]) k^2 C[
    1]^2)/(-1 + E^((x C[1])/2 + (C[1] C[2])/2) m)^2}, {f[x] -> (
   E^(x C[1] + C[1] C[2]) k^2 C[
    1]^2)/(1 + E^((x C[1])/2 + (C[1] C[2])/2) m)^2}}

In order to define functions it is better to use Part like so:

a1[x_] = 
 f[x] /. DSolve[{m f[x]^(3/2) f'[x] + 2 k f'[x]^2 - 2 k f[x] f''[x] ==
       0}, f[x], x][[1]]
a2[x_] = 
 f[x] /. DSolve[{m f[x]^(3/2) f'[x] + 2 k f'[x]^2 - 2 k f[x] f''[x] ==
       0}, f[x], x][[2]]

Now we can move on and ask Mathematica to Solve

Solve[a[0] == 1, C[1]]

and it complains a bit

This system cannot be solved with the methods available to Solve.

We can use assumptions to get

Assuming[k != 0, Solve[(a1[0] /. C[2] -> 0) == 1, C[1]]]

{{C[1] -> (1 - m)/k}, {C[1] -> (-1 + m)/k}}

Without assumptions, further progress can be made with the use of Reduce

Reduce[a1[0] == 1, C[1]]

giving

(-1 + m != 0 && k != 0 && C[2] == 0 && (C[1] == (-1 + m)/k || C[1] == (1 - m)/k)) || (C[ 3] \[Element] Integers && C[2] != 0 && -1 + E^((C[1] C[2])/2) m != 0 && k != 0 && (C[1] == ( m C[2] + 2 k ProductLog[C[3], -((E^(-((m C[2])/(2 k))) C[2])/(2 k))])/( k C[2]) || C[1] == (-m C[2] + 2 k ProductLog[C[3], (E^((m C[2])/(2 k)) C[2])/(2 k)])/( k C[2])))

Answer 2

The following goes some way towards a solution, by using Reduce, and by including all parameters in the solution in the list of variables to be solved for:

sol = DSolve[m f[x]^(3/2) f'[x] + 2 k f'[x]^2 - 2 k f[x] f''[x] == 0, f[x], x]
Reduce[f[x] == 1 /. sol /. x -> 0, {C[1], C[2], k, m}]

(* Out:
C[1] != 0 && (k == -(1/(Sqrt[E^(C[1]*C[2])]*C[1])) || 
   k == 1/(Sqrt[E^(C[1]*C[2])]*C[1])) && m == 0
*)