# System of equations

jaredmt

## Homework Statement

solve for x and y:

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

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

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

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

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)

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.

jaredmt
thanks but i don't think that's 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 don't understand the theory behind it but i checked it with both equations and it works, so I am happy

Last edited:
Mentor
thanks but i don't think that's 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
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