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anonymous
 5 years ago
Find the general solution to the following system:
dx/dt= x+y
dy/dt=2y
anonymous
 5 years ago
Find the general solution to the following system: dx/dt= x+y dy/dt=2y

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so solving the 2nd equation first I get \[y=ce ^{2t}\] Then work for the second one: \[dx/dt = x + ce ^{2t}\] \[x = \alpha ce ^{2t}\] \[x' = 2\alpha ce ^{2t}\] \[2\alpha ce ^{2t}\alpha ce ^{2t} = ce ^{2t}\] Therefore \[\alpha = 1\] and \[x=ce ^{2t}\] and the general solution is (ce^2t,ce^2t). But I think that's wrong. Anyone shed some light?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There is one solution missing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In the second equation, you only found the particular solution.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let me explain what I mean. From the 2nd equation, you found that: \[y=ce^{2x}\] And then you substitute that into the first equation, and that gives you: \[x'=x+ce^{2t} \implies x'x=ce^{2t}\] You found the particular solution for this which is: \[x_p(t)=ce^{2t}\] The complimentary solution can be found using the auxiliary equation, associated with the homogeneous DE: \[m1=0 \implies m=1 \implies x_c(t)=ke^t\] Therefore the general solutions are:\[x(t)=ce^{2t}+ke^t\]\[y(t)=ce^{2t}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does that makes sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, I think so. I'm going to try to find that section of my book and review more as well. I appreciate the help .
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