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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!

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1 Answer 1

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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

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

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