## 3rd-order to first order ODE and solving

Hi
I have an initial value problem to solve.
It is a third-order ODE that must be converted to first-order ODE's first.

$\displaystyle 2\frac{d^3f}{dn^3} + f\frac{d^2f}{dn^2} = 0$

Where:
$\displaystyle f(0) = 0$
$\displaystyle \frac{df}{dn}(0) = 0$
$\displaystyle \frac{df}{dn}(\infty) = 1$
I took variables as such:

$\displaystyle x1 = f$
$\displaystyle x2 = \dot{f}$
$\displaystyle x3 = \ddot{f}$

This gives me the first order equations:

$\displaystyle \dot{x1} = x2$
$\displaystyle \dot{x2} = x3$
$\displaystyle \dot{x3} = -\frac{x1x3}{2}$

Have been told that $\displaystyle \infty$ can be presumed at 10
Also that I have to find a solution for $\displaystyle \frac{d^2f}{dn^2}(0) = ?$ which should lie in the range [0.1,0.4] such that $\displaystyle \frac{df}{dn}(10) = 1$

So the idea is to find a value for $\displaystyle \dot{x2}(10)$ as this is $\displaystyle \frac{d^2f}{dn^2}(10)$ and then use that to see whether $\displaystyle \frac{df}{dn}(10) = 1$ or not.

So...
I used Euler's method and MATLAB to iterate up to (10) as follows,

Code:
x3(i+1)= x3(i) + x3(i)*h;
giving me the value $\displaystyle \dot{x2}$.

I have no idea how I'm supposed to continue past this point. I get the idea of using the bisection method to home in on the correct initial condition so that $\displaystyle \frac{df}{dn}(10) = 1$ as the first guess will of course be wrong, but getting to that I'm just stumped.

Any help gratefully appreciated,
Matt