# From a System of 1st ODE to a 2nd ODE and back to the system of 1st ODEs

Bachelier
I have a somewhat theoretical question regarding Differential Equations:

How can we reconcile the fact that if I go from let's say this system of 1st ODE

x' = 2y-x
y' = -x+y

to a 2nd ODE "using x(t) instead of y(t)" we get: x" + x =0

then back to a system of 1st ODE:

letting y=x' and then from x"=-x we get x"=-y' so y'=-x.

So now our new 1st ODE system has the following equations instead

x'=y
y'=-x

which is different from our original system:

x' = 2y-x
y' = -x+y

Is there a good Mathematical explanation to this?

Homework Helper
I have a somewhat theoretical question regarding Differential Equations:

How can we reconcile the fact that if I go from let's say this system of 1st ODE

x' = 2y-x
y' = -x+y

to a 2nd ODE "using x(t) instead of y(t)" we get: x" + x =0
I don't know what you mean by "using x(t) instead of y(t)". And the x satisfying the system of equations definitely does NOT satisfy x"= x= 0. What we can do is differentiate the first equation again to get x"= 2y'- x. Then, since y'= -x+ y, x"= 2(-x+ y)+-x= -3x+ 2y. Since, from the first equatioj, 2y= x'+ x, x"= -3x+ (x'+ x)= x'- 2x or x"- x'+ x= 0.

then back to a system of 1st ODE:

letting y=x' and then from x"=-x we get x"=-y' so y'=-x.

So now our new 1st ODE system has the following equations instead

x'=y
y'=-x

which is different from our original system:

x' = 2y-x
y' = -x+y

Is there a good Mathematical explanation to this?
However, your question is still valid.

If, in the differential equation x"- x'+ x= 0, you let y= x', then x"= y' and so you have y'- y+ x= 0 or y'= y- x.

Now the system is

x'= y,
y'= y- x.

Yes, that is different from the original system because this "y" is different from the y in the original system.