2nd-order ODEs

One-step methods can be used to solve higher order equations. This is similar to the substitution method for analytical ODE solutions covered in MTE 202.

Let’s say we have a 2nd order IVP:

$$\begin{array}{r}\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y=e^{-x}\\ y(0)=1,\frac{dy(0)}{dx}=0\end{array}$$

Then, we let $dy/dx=v$ and substitute, so that we have:

$$\frac{dv}{dx}+xv+y=e^{-x}$$

And the IC becomes:

$$y(0)=1,v(0)=0$$

We can solve this as a first-order problem $h=0.5$:

$$\begin{array}{rl}{y}_{i+1}& =y_i+v_{i}h\\ v_{i+1}& =v_i+[e^{-x_i}-x_i{v}_i-y_i]h\end{array}$$

Start at $i=0,x_0=0,y_0=1,v_0=0$:

$$\begin{array}{rl}{y}_1& =1+0.5(0)=1\\ v_1& =0+0.5(e^0-0-1)=0\end{array}$$