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October 22nd, 2015, 12:15 PM   #1
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Can someone please explain this IVP is solved?

dx/dt = -5x - y
dy/dt = 4x - y

I got
x = ae^-3t + bte^-3t
y = -2bte^-3t + 2ae^-3t + be^-3t
a = 0, b = 0

The answer is
x = e^(-3t+3) - te^(-3t+3)
y = -e^(-3+t) + 2te^(-3t+3)
I don't understand where the +3 comes from.

Last edited by skipjack; October 28th, 2015 at 07:22 AM. Reason: fix typos
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October 22nd, 2015, 01:10 PM   #2
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It's not possible to determine what the original problem was from your post with any certainty.
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October 28th, 2015, 05:05 AM   #3
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Quote:
Originally Posted by marchingt9 View Post
x =-5x-y
y =4x-y
You mean, surely, x'= -5x- y, y'= 4x- y?
And this is NOT an "initial value problem" because you haven't stated any initial values!

Quote:
I got
x=ae^-3t+bte^-3t
y=-2bte^-3t+2ae^-3t+be^-3t
a=0 b=0

The answer is
x=e^(-3t+3)-te^(-3+3)
y=-e^(-3+t)+2te^(-3+3)
I dont understand where the +3 comes from
From your as yet unstated initial values! Apparently your coefficients, a and b, are e^3.
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October 28th, 2015, 08:14 AM   #4
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As per Country Boy's observations, I've corrected some typos in the original post, except for the -e^(-3+t) term, which presumably should have been -e^(-3t+3). I'm guessing that the initial values of x and y were given as e³ and -e³.

The "+3" in the exponents is then equivalent to a factor of e³. Clearly, marchingt9's "a = 0, b = 0" is incorrect, but I'll leave marchingt9 to correct that and anything else that needs correcting.

One method of solving the system starts by adding double the first equation to the second equation. The resulting equation is easily solved to give 2x + y in terms of t.
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