# Second order linear differential equation with constant coefficients

1. Jan 25, 2012

### stripes

1. The problem statement, all variables and given/known data

For the differential equation y'' - 4y' + y = 0,
(a) Show that if we let x = y' (i.e. x(t) = y'(t)), then this leads to the system:
x' = 4x -y
y' = x

(b) Conversely, show that the system in (a) leads to y'' - 4y' + y = 0 (and x'' -4x' + x = 0 also).

2. Relevant equations

None.

3. The attempt at a solution

part (a) seems easy enough, let x = y', then x' must equal y'', then substitute:

y'' - 4y' + y = 0
x' - 4x + y = 0
x' = 4x - y as expected (and then we have x = y' initially given), so the system is correct.

part (b) seems extremely obvious, which makes me wonder if I'm missing something. Anyways,, I begin with

x' = 4x - y and y' = x. y'' = x' follows.

by substituting back in, we have
y'' = 4y' - y
y'' - 4y' + y = 0 as required.

To show find x'' - 4x' + x = 0, I don't know. I tried finding dy/dx:

dy/dx = y'/x' = x/(4x - y) which is not separable:
dy(4x - y) = xdx

so my questions are: is part (b) as simple as it seems? Or is my method not correct? And how do I show x'' - 4x' + x = 0? Thanks in advance!

2. Jan 25, 2012

### vela

Staff Emeritus
Yes, it's as simple as it seems.

You find the differential equation for x similarly. Start by differentiating the equation for x'.

3. Jan 25, 2012

### stripes

Thank you for your help. I end up with x'' - 4x' + x = 0 as required.