Question

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

Answer 1

is it a matter of setting up a dy and y-terms on one side and the dt and t-terms on the other?

Answer 2

Do you know about the integrating factor?

Answer 3

vaguely, but i'm not too sure on it yet

Answer 4

For differential equation of the type
\(y'+p(t)y=q(t)\)
the trick is to multiply the equation by \(e^{\int p(t)\, dt}\)

Answer 5

so in this case, \[e ^{\int\limits_{}^{} e ^{3t}dt}\] ?

Answer 6

yes :)

Answer 7

well to be precise since you move the \(e^{3t}y\) to the left, your integrating factor is
\(e^{\int -e^{3t}\,dt}\)

Answer 8

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.

Answer 9

how would that be done with the y' factor?

Answer 10

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.

Answer 11

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.

Answer 12

hmm what is derivative of \(e^{-e^{3t}/3}y\) with respect to \(t\)?

Answer 13

\[e ^{-e ^{3t}}y\] ?