# Homework Help: System of linear diff eqs

1. Oct 7, 2008

### LMKIYHAQ

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

Consider the system x'= -x+y
y'= -2y
Find the general solution (x(t),y(t)) by first solving for y(t) and then solving for x(t).

2. Relevant equations

Should I use undetermined coefficients for each of these separate diff eqs or should I substitute x'+x into the y' +2y=0?

3. The attempt at a solution
When I attempted this, I substituted x'+x into y'+2y=0 and got ae^-2t +be^-t where a and b are arbitraty constants. I am not sure if I need to look at a particular solution using undetermined coefficients or if I am even on the right track??

2. Oct 7, 2008

### gabbagabbahey

Well, the problem tells you to solve y'=-2y first, so what is the general solution y(t) to this homogeneous equation?

3. Oct 7, 2008

### LMKIYHAQ

I solved this and got y(t)= -(e^(t+c))/2. Is this right?

4. Oct 7, 2008

### gabbagabbahey

Well,

$$y'(t)=\frac{d}{dt}\left( \frac{-1}{2} e^{t+c}\right)= \frac{-1}{2} e^{t+c} \neq -2y$$

so I'm gunna say no on that one.

The DE y'(t)=-2y(t) tells you that the time derivative of y(t) is proportional to y(t), and that the constant of proportionality is -2. I can think of only one type of function that could account for that: Ae^(-2t)

Last edited: Oct 7, 2008
5. Oct 7, 2008

### LMKIYHAQ

ok so y(t)=ae^(-2t). I think I understand how to get this by using undetermined coeff. where y'+2y=0.
For x(t)... x'(t)= -x+y...what information do I have for the y part in order to slove for x(t)?

6. Oct 7, 2008

### gabbagabbahey

You have y=ae^(-2t); plug that in and use it:

$$x'=-x+ae^{-2t} \Rightarrow x'+x=ae^{-2t}$$

which is an inhomogeneous DE which you can solve using undetermined coefficients. Start with the complimentary solution $x_c(t)$, what do you get for that?