Question:
Suppose that $f$ satisfies
$$f''' + \frac 12 ff''=0$$
with $f(0) = f'(0) = 0$ and $f(x) \sim x$ as $x \rightarrow \infty$. Find the value of $f''(0)$ to leading order.
Attempt:
The question does say to find $f''(0)$ only to leading order, and indeed no matter how hard I tried I couldn't find an exact solution or even a first integral to the above ODE.
Then, I tried a Frobenius series, expanding about the point $x=0$:
$$f(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2+a_3x^3+\cdots$$
The conditions $f(0) = f'(0) = 0$ imply that $a_0 = a_1 = 0$, so
$$f(x) = a_2x^2+a_3x^3+a_4x^4+\cdots$$
Plugging this in, we find that
$$\big(60a_5 + a_2^2 \big)x^2 + \mathcal O(x^3) = 0$$
We can of course continue expanding, but the problem is that there is no way to use the final condition at $x = \infty$ because this expansion is only valid near $x=0$.
And then I ran out of ideas.
Any hints? Thanks!
Hint:
With reference to http://eqworld.ipmnet.ru/en/solutions/ode/ode0503.pdf,
Let $u=\left(\dfrac{df}{dx}\right)^2$ ,
Then $\dfrac{du}{dx}=2\dfrac{df}{dx}\dfrac{d^2f}{dx^2}$
$\dfrac{du}{df}\dfrac{df}{dx}=2\dfrac{df}{dx}\dfrac{d^2f}{dx^2}$
$2\dfrac{d^2f}{dx^2}=\dfrac{du}{df}$
$2\dfrac{d^3f}{dx^3}=\dfrac{d}{dx}\left(\dfrac{du}{df}\right)=\dfrac{d}{df}\left(\dfrac{du}{df}\right)\dfrac{df}{dx}=\pm\sqrt u\dfrac{d^2u}{df^2}$
$\therefore\pm2\sqrt u\dfrac{d^2u}{df^2}+f\dfrac{du}{df}=0$
Which reduces to a generalized Emden-Fowler equation