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Initial Value Problem

  • Thread starter Nayr
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  • #1
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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...
 
Last edited:

Answers and Replies

  • #2
Dick
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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...
That, in fact, is perfectly ok. For the other side think about doing a substitution.
 
  • #3
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Ah, I see. Thanks for clearing that up. Now it's the x side that's cleared up and the y side that I'm stuck on. I'll go through the substitution loops and see what I get...

Hm. Doesn't really seem like I can get a clean substitution... Gonna keep at it, though.

I think I have it. Ended up with y = ln(1+cos(x)) + C, with C = ln(2)
 
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  • #4
HallsofIvy
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To integrate sin(x)/(cos(x)+1) dx, let u= cos(x)+1, so that du= -sin(x)dx.

To integrate 1/(e-x+ 1) dx, let u= e-x+ 1, so that du= -e-x dx. e-x= u- 1: du= -(u-1)dx so dx= -1/(u-1) du. 1/(e-x+ 1)dx= 1/(u(u-1)) du.
 
  • #5
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Thank you very much! The substitutions really help to put things into perspective.

However... I am now having an embarrassing amount of trouble integrating 1/(u(u-1)) du.

I think it comes out to ln(u/u+1)), but when I try to expand and then unsubstitute, one of my variables cancels itself out.

The answer given in the back of the book is (1 + cos(x))(e^(y) + 1) = 4. I tried working backwards from there, but to no avail. What I'm not sure of is how the exponential y becomes positive while staying grouped with the +1. Not sure if what I just said makes sense, but after spending 6 nonconsecutive hours on this one problem today, my brains are turning to mush...
 
  • #6
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However... I am now having trouble integrating 1/(u(u-1)) du.
Break it into partial fractions,

[tex]\frac{A}{u} + \frac{B}{u-1} [/tex]
 
  • #7
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Aaaah! I knew that there was something that I'd forgotten! I haven't used partial fractions since last year, so it just slipped my mind... Thank you so much!
 

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