dy/dx=(cosx)e^(y+sinx)
instructions say to 'use separation of variables to solve the initial value problem'
answer is y=ln(2-e^(sinx))
I get to y=-ln(sinx(e^sinx))+c without trouble, but I'm not sure where the 2 comes from or how to get rid of that extra sinx.

Question

tkhunny · Answer

Let's see...

$\dfrac{dy}{dx} = \cos(x)e^{y+\sin(x)} = e^{y}\cos(x)e^{\sin(x)}$

$\int e^{-y}dy = \int \cos(x)e^{\sin(x)}dx$

$-e^{-y} = e^{\sin(x)} + C$

That's odd.  $e^{-y} > 0$, so logarithms are not appropriate.

$e^{-y} = -e^{\sin(x)} + D = D - e^{\sin(x)}$

Now, we can introduce the logarithm

$ -y = \ln \left( D - e^{\sin(x)} \right) $

$ y = -\ln \left( D - e^{\sin(x)} \right) $

Looks to me like you're struggling with three things:
1) Just what is it with that Arbitrary Constant?  What does it do?  How is it used?
2) Some algebra issues.
3) Maybe the trigonometric antiderivative?

I think we just need the initial conditions in order to get D = 2.

anonymous · Answer

Okay, thanks...one more question, in the third step, where does the cosx go? Wouldn't the antiderivative of that be sinx, so it would stick around?

Other than that, very helpful though. I definitely have some issues with algebra, I'm kind of filling in the gaps as I go :)

tkhunny · Answer

See point #3?  $\dfrac{d}{dx}e^{\sin(x)} = e^{\sin(x)}\cdot \cos(x)$

You'll have to get up to speed on your chain rule for derivatives and your substitution methods for integrals.

anonymous · Answer

Okay, that could possibly have been very idiotic XD I get it now, thanks again!

tkhunny · Answer

No worries.  We're all learning.  Be careful and thorough.

Point #1 is also very important.  You just added "+ c" on the end after you were done.  That's no good.  It appears at the end of any indefinite integral.  After that, you have to keep track of it differently.

anonymous · Answer

So just making sure, I need to add C right after I integrate and then treat it as a variable?

tkhunny · Answer

Not a variable, but as something that is a little less arbitrary than it was before.  You can see that I changed from "C" to "D".  This wasn't really necessary, but it was clear to me that I was about to change the sign of "C".  Rather than write "-C" I decide to write "+D".  This was just to emphasize that it was a different constant.

A very common tranformatino of a constant is this one.

ln(y) = ln(x) + C

Applying Logarithms

$y = e^{ln(x)+C} = e^{ln(x)}e^{C} = Ae^{ln(x)} = Ax$

We had "C", but after the logarithm, we know it is a different constant, but we really don't know what it is, so we just renamed it.

anonymous · Answer

Okay, cool. Thanks!