# Get an analytical solution of a non-linear 3rd order ode?

Posted 6 months ago
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 The equation that I need an analytical solution of, is $f'''= \frac{1}{3}f'^2+\frac{2}{3}ff''$The boundary conditions are:at $x=0; f=0, f'=1$at $x=\infty; f'=0$If I use Dsolve, it doesn't return a solution sol = Dsolve[y'''[x] == (1/3)*(y'[x])^2 - (2/3)*(y[x])*(y''[x]), y[0] == 0, y'[0] = 1, y'[inf] == 0, y[x], x] 
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Posted 6 months ago
 Likely, that means Mathematica doesn't know a closed-form solution, if it exists.  DSolve[{f'''[x] == 1/3*f'[x]^2 - 2/3*f[x]*f''[x], f[0] == 0, f'[0] == 1, f'[Infinity] == 0}, f[x], x] (* Returns unevaluated *) Numeric method: sol = NDSolve[{f'''[x] == 1/3*f'[x]^2 - 2/3*f[x]*f''[x], f[0] == 0, f'[0] == 1, f'[20] == 0}, f, {x, 0, 20}] Plot[{f[x] /. sol, Evaluate@D[f[x] /. sol, x]}, {x, 0, 20}, PlotLegends -> {"f[x]", "f'[x]"}] 
Posted 6 months ago
 Thanks, the numerical solution is what I expected it to be.
 It can be reduced to second-order ODE with substitution: $f'''(x)=u''(f) u(f)^2+u'(f)^2 u(f)$ $f''(x)=u(f) u'(f)$ $f'(x)=u(f)$ where $f(x)=f$Putting to equation:  u''[f]*u[f]^2 + u'[f]^2*u[f] == 1/3*u[f]^2 - 2/3*f*u[f]*u'[f] u''[f]*u[f] + u'[f]^2 == 1/3*u[f] - 2/3*f*u'[f] eq2 = u''[f]*u[f] + u'[f]^2 + 2/3*f*u'[f] - 1/3*u[f] == 0 DSolve[eq2, u[f], f] (* Returns unevaluated *) With help of Maple it can be reduced to Abel second kind equation,but DSolve returns unevaluated.  DSolve[y'[x]*y[x] - x^2/3 + 2/3*y[x] == 0, y[x], x] ( * Input ?*) I can only say Mathematica 11.3 and Maple 2018 doesn't know a closed-form solution.