# System of equations

1. Apr 14, 2009

### jaredmt

1. The problem statement, all variables and given/known data
solve for x and y:

tx' = x + y
ty' = -3x + 5y

2. Relevant equations

3. The attempt at a solution

normally if the t was not in front of the primes then i'd set it up as:
(D-1)x -y = 0
(D+3)x -5y=0

but im not sure what to do when the variable t is there. can somebody help me set this up?

2. Apr 14, 2009

### whybother

Sorry, just to be clear, are there 5 variables in this problem? t, x', y', x, y, or are some of those constants?

3. Apr 15, 2009

### jaredmt

yes but x' is dx/dt and y' is dy/dt

4. Apr 18, 2009

### jaredmt

nobody knows how to do this? i'll put it in a different form if it makes more sense this way:

t*(dx/dt)= x + y
t*(dy/dt) = -3x + 5y

then solve for y(t) and x(t)

5. Apr 18, 2009

### Billy Bob

Would this work?

dx/(x+y)=dt/t

and

dy/(-3x+5y) = dt/t.

Therefore set

dx/(x+y)=dy/(-3x+5y)

and solve for y=y(x) or x=x(y). I can do that by using y=vx. Not sure what would come next, though.

6. Apr 18, 2009

### jaredmt

thanks but i dont think thats what they're looking for.

in the book it says set t = e^w and it should become a linear system with constant coefficients.
then i would get:

x'e^w = x + y
y'e^w = -3x + 5y

can i treat e^w as a constant and solve the problem then replace w with ln(t) ?

i guess i'll try this and see what happens

edit: ok i figured out how to get the right answer. i dont understand the theory behind it but i checked it with both equations and it works, so im happy

Last edited: Apr 18, 2009
7. Apr 18, 2009

### Staff: Mentor

Since t = e^w, you can't treat e^w as a constant if t isn't a constant.
The idea of this substitution is to get rid of the t factors on the left sides.

t = e^w ==> dt = e^w * dw
dx/dt * e^w = x + y
dy/dt * e^w = -3x + 5y

or
dx/(e^w * dw) * e^w = x + y
dy/(e^w * dw) * e^w = -3x + 5y

On the left sides, the e^w factors cancel, so you get
dx/dw = x + y
dy/dw = -3x + 5y

Presumably you can solve this system of linear first-order DEs.