- #1

Nayr

- 4

- 0

I need some help with an initial value problem. It looked pretty easy at a glance, but I've ended up spending over 3 hours on this one problem.

The problem is as follows:

(e^(-y)+1)(sin(x) = (1+cos(x)) dy/dx

With y(0) = 0.

I rearranged the equation to get

sin(x) / (1+cos(x)) dx = 1 / (e^(-y)+1) dy

But at this point, I can't figure out how to integrate the two equations. I'm fairly (but not entirely) sure that the antiderivative of the y side is simply ln(e^(y)+1), but I'm at a loss for the x side. Is there something simple I'm missing, or did I go about the entire problem incorrectly?

Edit:

Moments after posting this, I realized that the antiderivative of sin(x) / (cos(x) +1) is (cos(x)+1)^-1+C. I'll see where I can go from there and edit if I get anything out of it.

Re-Edit:

Nope, actually, that was pretty dumb. Just realized that the 1+cos(x) would have to be squared under the denominator for that to be true. Back to square 1! Perhaps it's -ln|1+cos(x)|? But I think you can't do that with trig functions...

The problem is as follows:

(e^(-y)+1)(sin(x) = (1+cos(x)) dy/dx

With y(0) = 0.

I rearranged the equation to get

sin(x) / (1+cos(x)) dx = 1 / (e^(-y)+1) dy

But at this point, I can't figure out how to integrate the two equations. I'm fairly (but not entirely) sure that the antiderivative of the y side is simply ln(e^(y)+1), but I'm at a loss for the x side. Is there something simple I'm missing, or did I go about the entire problem incorrectly?

Edit:

Moments after posting this, I realized that the antiderivative of sin(x) / (cos(x) +1) is (cos(x)+1)^-1+C. I'll see where I can go from there and edit if I get anything out of it.

Re-Edit:

Nope, actually, that was pretty dumb. Just realized that the 1+cos(x) would have to be squared under the denominator for that to be true. Back to square 1! Perhaps it's -ln|1+cos(x)|? But I think you can't do that with trig functions...

Last edited: