Question

find general solution dy/dt-y=3e^t

Answer 1

Cash moneyyyyyyyy.
So what's up? Integrating factor, yes?

Answer 2

yes

Answer 3

\[\Large\rm y'-y=3e^t\]We want to multiply through by some integrating factor,\[\Large\rm \mu y'-\mu y=3\mu e^t\]So that the left side will simplify, applying product rule in reverse,\[\Large\rm (\mu y)'=3\mu e^t\]And then it's not too bad from there.

Answer 4

So we have the form:\[\Large\rm y'+p(t)y=f(t)\]Our integrating factor will be,\[\Large\rm \mu =e^{\int\limits p(t)dt}\]So in this case our p(t) is -1, yes?

Answer 5

Confused by any of that crazy stuff?

What do you get for your \(\Large\rm \mu\) ?

Answer 6

uhhh....e^-t

Answer 7

Good good good, that's what we're multiplying through by.\[\Large\rm e^{-t}y'-e^{-t}y=3e^{-t}e^{t}\]Left side and right side will simplify down,\[\Large\rm (e^{-t}y)'=3\]Confused by any of that?

Answer 8

After that you just integrate! :O

Answer 9

how did it get simplify to (e^-t(y))'=3

Answer 10

On the right side, exponent rule,
\[\Large\rm e^{-t}e^{t}=e^{-t+t}=e^0=1\]

Answer 11

The left side collapses because of the product rule, but in reverse. That is actually the purpose of our integrating factor: it makes the side with the function and its derivative into a product rule form and allows you to simplify it that way!

So if you worked backwards, you could expand this out using product rule:
\( (e^{-t} y)' = e^{-t} (y)' + (e^{-t})' y = e^{-t} y' - e^{-t} y' \)
And you are back at your original left side. Does this all make sense?

Answer 12

how would integrate it?

Answer 13

@AccessDenied @zepdrix