1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integrating Factor Method

  1. Sep 12, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the general solution to the indicated equation:
    [itex] cos(x)y' + ysin(x) = 1 [/itex]

    2. Relevant equations
    [itex] e^\int p(x)\,dx * y(x) = int\ f(x)\, dx e^\intp(x)\,dx + C [/itex]

    3. The attempt at a solution
    Ok, I am having trouble getting started with this problem because I am not really sure what value to assign to p(x). I attempted to assign p(x) = sin(x) therefore u(x) = e^cosx. However, I do not believe that e^cosx can be integrated without using some advanced techniques, so I am quite sure I am not doing the problem correctly.
    Edit: when I say e^cosx cannot be integrated I mean not in terms of elementary functions
    Edit: Sorry about the bad Latex equations, I am attempting to sort it out (I have never used it before)
    Last edited: Sep 12, 2014
  2. jcsd
  3. Sep 12, 2014 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    To calculate the integrating factor your equation needs to be in the form$$y' + p(x)y = q(x)$$So you need to start by dividing your equation through by ##\cos x## before trying to evaluate the integrating factor.
  4. Sep 12, 2014 #3
    thank you, so then the equation would become:
    [itex] y' + ytan(x) = sec(x) [/itex] and
    [itex] p(x) = tan(x) [/itex]
  5. Sep 12, 2014 #4
    Yes. The integrating factor of [itex]e^{\int \tan x \,dx}[/itex] should now work as advertised. :) Just be sure to note whether the points for which cos(x) = 0 are to be included in the domain of the solutions.
  6. Sep 12, 2014 #5
    Edit: I mean can [itex] e^{sec^2x} * secx [/itex] be integrated? Because that is what I'll end up with no?
  7. Sep 12, 2014 #6
    sec2(x) is not an integral of tan(x). Remember we need to find a function that has tan(x) as its derivative. Try writing tan(x) as [itex]\frac{\sin x}{\cos x}[/itex].
  8. Sep 12, 2014 #7
    ahh...thanks, sorry about the silly mistake
  9. Sep 12, 2014 #8
    ok, for my final answer I am getting:
    [itex] y(x) = sin(x) + \frac {C}{sec(x)} [/itex]
  10. Sep 12, 2014 #9
    That looks fine. However, it may be preferable to write the solution as [itex]y(x) = \sin x + C\cos x[/itex], since both this function and the original differential equation is defined for all x, while your function is not defined for some of the values of x for which the original differential equation is defined. :)
  11. Sep 12, 2014 #10
    thanks again, I'll do that
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted