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!