Solve (differential equations): dy/dt = y*e^3t + e^-t

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Solve (differential equations): dy/dt = y*e^3t + e^-t

Mathematics
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is it a matter of setting up a dy and y-terms on one side and the dt and t-terms on the other?
Do you know about the integrating factor?
vaguely, but i'm not too sure on it yet

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Other answers:

For differential equation of the type \(y'+p(t)y=q(t)\) the trick is to multiply the equation by \(e^{\int p(t)\, dt}\)
so in this case, \[e ^{\int\limits_{}^{} e ^{3t}dt}\] ?
yes :)
well to be precise since you move the \(e^{3t}y\) to the left, your integrating factor is \(e^{\int -e^{3t}\,dt}\)
So that integrating factor is \(e^{-e^{3t}/3}\). Multiply the equation by that we have \(e^{-e^{3t}/3}y'-e^{-e^{3t}/3}e^{3t}y=e^{-e^{3t}/3}e^{-t}\) \((e^{-e^{3t}/3}y)'=e^{-t-e^{3t}/3}\) then integrate both sides.
how would that be done with the y' factor?
Notice that if we apply the the product rule to compute the derivative of \(e^{-e^{3t}/3}y\) we will get the previous step.
how would all that simplify? i'm not sure if i'm mixing up rules or what, but i'm coming up with far-out answers that don't simplify.
hmm what is derivative of \(e^{-e^{3t}/3}y\) with respect to \(t\)?
\[e ^{-e ^{3t}}y\] ?

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