Why is Mathematica struggling to solve this ODE with a known analytic solution?

[member 428835]

Hi PF!

I am wondering if any of you can explain why Mathematica won't solve this ODE? I know an analytic solution exists (it's quadratic). Any help is greatly appreciated! $f0[x]$ is the name of the function.

a = -1/5;
DSolve[{-((2 a f0[x])/(1 + a)) + x f0'[x] + 
    2 (f0'[x])^2 + f0[x] f0''[x] == 0, 
  f0[1] == 0, f0'[1] == -1/2}, f0[x], x]

 

[JorisL]

Some hints to improve your post.
Give us the exact differential equation in LaTeX. having to "translate" hardly readable mathematica code doesn't help getting answers.
You say you know an analytic solution exists, do you know the form as well?

More on topic, have you tried NDSolve to see if mathematica can work with the DE and gives a reasonable result?

 

[member 428835]

$$f f''+2f'^2+x f'+\frac{1}{2}f = 0$$ where $f$ is a function of $x$ subject to $f(1)=0$ and $f'(1)=-1/2$. The analytic solution is $f=(1-x^2)/2$.

I have tried NDSolve and it works great, provided I perturb the boundary condition as $f(1)=0.00001$.

Any ideas why it can't figure out the solution?

 

[JorisL]

So I've been looking at this for a bit.

First I would get rid of the factor of $f$ from the second derivative term.
From that I would be led to believe you could consider $g = f^\prime/\sqrt{f}$ as a new variable.

$g^\prime = \frac{f^{\prime\prime}}{\sqrt{f}} - \frac{f^\prime}{f}$
Then $f^{\prime\prime} = \left(g^\prime+\frac{f^\prime}{f}\right)\sqrt{f}$
So we can rewrite your DE as

$\sqrt{f}\left(g^\prime + g\right) + 2 \left(g^\prime\right)^2+xg+\frac{1}{2} = 0$

That's where I got stuck, maybe this gives you some ideas.
I think the key is rewriting the equation such that mathematica recognizes it.
Also wolfram alpha gave me a solution (for different initial values) although no analytical expression.
Since wolfram alpha uses mathematica in the back this is promising.

 

[Daeho Ro]

Are u sure the solution of that differential equation is really $f=(1-x^2)/2$? I didn't get it because the initial condition of this solution is not same as you mentioned and actually it is $f'(1)=-1$.

I think the factor $1/2$ is missing. However, still I can't get the solution same as you. That's interesting.