# Using Implicit Euler with second order differential equations

We can numerically integrate first order differential equations using Euler method like this: $$y_{n+1} = y_n + hf(t_n, y_n)$$

And with Implicit Euler like this: $$y_{n+1} = y_n + hf(t_{n+1},y _{n+1})$$

If I have a differential equation $$y' - ky = 0$$, I can integrate $$y$$ numerically using Implicit Euler: $$y_{n+1} = y_n + hky_{n+1}$$ $$y_{n+1} = y_n\frac{1}{1-hk}$$

But how I do use Implicit Euler for second order differential equations, like for instance the equation for simple harmonic motion? $$y'' + w^2y = 0$$

We have to integrate with respect to $$y$$ and $$y'$$. For explicit Euler the numerical integration would look like this (?): $$y_{n+1} = y_n + hf(t_n, y'_n)$$ $$y'_{n+1} = y'_n + hg(t_n, y_n)$$

How would we do integrate using Implicit Euler instead?

• It is almost always a good idea to express higher-order ODEs in their equivalent first-order form. – pluton Nov 13 '20 at 17:05

Yes. For convenience, name the derivative $$v_n=y_n'$$. Then \begin{align} y_{n+1}&=y_n+hv_{n+1}\\ v_{n+1}&=v_n-hw^2y_{n+1}\\ \implies y_{n+1}(1+h^2w^2)&=y_n+hv_n\\ v_{n+1}(1+h^2w^2)&=v_n-hw^2y_n \end{align} which looks like an explicit Euler method with a correction factor.