# How to turn this second order ODE into a system of two first order ODEs?

I was given this problem in a homework set and I don't know where to start -- haven't done diff equation in a while, so any help would be very appreciated!!

For small fluctuations of light, the photo-receptor can be considered linear and the equation describing its response is:

$$\frac{d^2x}{dt^2}+1.5\frac{dx}{dt}+0.5~x(t)=6s$$

where $$s$$ is the light intensity (millenniums/mm$$^2$$) and $$x(t)$$ is the firing rate of the photo-receptor (Hz). Note that $$s$$ and $$x$$ are measured with respect to their nominal values, so $$s = 0$$ corresponds to normal illumination and the firing rate $$x = 0$$ is the deviation from some non-zero firing rate corresponding to normal illumination.

How do I convert this second order ODE into a system of two first order ODEs by defining the second state variable as $$y(t)=\frac{dx}{dt}$$ (rate of change of firing rate $$x(t)~$$?

Any advice or explanation would be very appreciated!!! Thank you!

It's pretty easy. Given

$$\ddot x + 1.5 \dot x + 0.5x = 6s, \tag 1$$

set

$$y = \dot x; \tag 2$$

then

$$\dot y = \ddot x; \tag 3$$

using (3), (1) may be written

$$\dot y = -1.5 y - 0.5 x + 6s; \tag 4$$

if we write (2) as

$$\dot x = y, \tag 5$$

then (4) and (5) together form a first order system in $$x$$ and $$y$$.

Of course, and equation such as (1) is typically supplied with initial conditions $$x(t_0)$$, $$\dot x(t_0)$$; when we transform the system, we take

$$y(t_0) = \dot x(t_0). \tag 6$$