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anonymous
 5 years ago
Solve (differential equations): dy/dt = y*e^3t + e^t
anonymous
 5 years ago
Solve (differential equations): dy/dt = y*e^3t + e^t

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is it a matter of setting up a dy and yterms on one side and the dt and tterms on the other?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Do you know about the integrating factor?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0vaguely, but i'm not too sure on it yet

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0For differential equation of the type \(y'+p(t)y=q(t)\) the trick is to multiply the equation by \(e^{\int p(t)\, dt}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so in this case, \[e ^{\int\limits_{}^{} e ^{3t}dt}\] ?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0well to be precise since you move the \(e^{3t}y\) to the left, your integrating factor is \(e^{\int e^{3t}\,dt}\)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0So that integrating factor is \(e^{e^{3t}/3}\). Multiply the equation by that we have \(e^{e^{3t}/3}y'e^{e^{3t}/3}e^{3t}y=e^{e^{3t}/3}e^{t}\) \((e^{e^{3t}/3}y)'=e^{te^{3t}/3}\) then integrate both sides.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how would that be done with the y' factor?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Notice that if we apply the the product rule to compute the derivative of \(e^{e^{3t}/3}y\) we will get the previous step.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how would all that simplify? i'm not sure if i'm mixing up rules or what, but i'm coming up with farout answers that don't simplify.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0hmm what is derivative of \(e^{e^{3t}/3}y\) with respect to \(t\)?
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