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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

    LCKurtz

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    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]
    right?
     
  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
     
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