I need to solve the following third order (non-linear) ODE by numerical methods: \begin{equation}\tag{1} h^{3} \dfrac{d^3 h}{d x^3} = h-1. \end{equation}
By assumption, the solution should approach $ h=1$ when $ x\rightarrow -\infty $. Furthermore, if we linearize the equation above by introducing $ h_{0} = 1+\varepsilon(x) $, where $ \varepsilon(x)\ll 1 $, the equation reduces to $ \dfrac{d^3 \varepsilon}{d x^3} = \varepsilon,\, $ which has the solution $ \varepsilon(x) = A\exp(x)+\exp(-x/2) \left[ B\cos(\frac{\sqrt{3}}{2} x) + C\sin(\frac{\sqrt{3}}{2} x) \right] $.
This means that \begin{equation} h_{0}(x) = 1+A\exp(x)+\exp(-x/2) \left[ B\cos(\frac{\sqrt{3}}{2} x) + C\sin(\frac{\sqrt{3}}{2} x) \right] \end{equation} is a solution to the linearized problem. Now, to ensure that $ h_{0}(x) \rightarrow 1 $ for $ x\rightarrow -\infty $ we must require that $ B=C=0 $. Thus, our solution to the linearized problem simplifies to: \begin{equation} h_{0}(x) = 1+A\exp(x). \end{equation} Here is the actual problem: Use this choice as an initial condition for solving eq. (1) numerically (preferably using Matlab) and show that \begin{equation} h''(x)\rightarrow K \quad\text{for}\quad x\rightarrow \infty \end{equation} and estimate the constant $ K $ (numerically).
Can anybody help? I'm not strong in numerical methods.
Hint:
This belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0503.pdf.
Let $y=\left(\dfrac{dh}{dx}\right)^2$ ,
Then $\dfrac{d^2y}{dh^2}=\pm2(h^{-2}-h^{-3})y^{-\frac{1}{2}}$
You can consider as two members Emden-Fowler type nonlinear ODE and follow the method in http://www.sciencepubco.com/index.php/ijamr/article/download/723/628