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

1. anonymous

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

2. watchmath

Do you know about the integrating factor?

3. anonymous

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

4. watchmath

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

5. anonymous

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

6. watchmath

yes :)

7. watchmath

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

8. watchmath

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.

9. anonymous

how would that be done with the y' factor?

10. watchmath

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.

11. anonymous

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.

12. watchmath

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

13. anonymous

$e ^{-e ^{3t}}y$ ?