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 October 22nd, 2015, 12:15 PM #1 Newbie   Joined: Oct 2015 From: Georgia Posts: 1 Thanks: 0 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
 October 22nd, 2015, 01:10 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra It's not possible to determine what the original problem was from your post with any certainty.
October 28th, 2015, 05:05 AM   #3
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Quote:
 Originally Posted by marchingt9 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.

 October 28th, 2015, 08:14 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,835 Thanks: 2162 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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