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Linear 1st order differential equations

  1. Oct 16, 2009 #1
    1. The problem statement, all variables and given/known data
    dy/dx + ytanx = secx, y(pi) = 1


    2. Relevant equations

    I(x) = e^integral P(x)dx
    Integral Q(x)e^integral P(x)dx
    y=e^-integral P(x)dx (integral Q(x)e^integral P(x)dx + C)

    (sorry, it is a bit messy, I don tknow how to use the math symbols yet)
    3. The attempt at a solution
    dy/dx + ytanx = secx, where P(x) = tanx, Q(x) = secx

    ytanx - secx + dy/dx = 0

    I(x) = e^Integral tanx = e^-ln(cosx)

    Integral Q(x)e^integral P(x)dx = Integral (1/cosx) e^-ln(cosx) dx.

    ^^this is where I am stuck, either I did something wrong in the beginning, and/or it has to be integrated by parts. I only know how to do the "tabular table" method for parts, so if my calculations were right to that point, could somebody help me through the rest of the problem. thanks :(
     
  2. jcsd
  3. Oct 16, 2009 #2

    Dick

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    It looks confusing because you aren't simplifying the integrating factor. exp(-ln(cos(x))) can be written in a much simpler form using rules of exponents and logs. What is it?
     
  4. Oct 16, 2009 #3
    Nm, i know how to solve it now, thanks.

    Although, I am stuck on this problem now:

    dy/dx - y = 4e^x, with y(0) = 4.

    How do I solve this when there isn't a P(x) function next to y...or is there...
     
  5. Oct 16, 2009 #4

    Dick

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    There is. P(x)=(-1).
     
  6. Oct 16, 2009 #5
    For dy/dx - y = 4e^x, with y(0) = 4.

    I got answer: y = (4x + C)e^x

    The answer in back says 4e^x (x + 1)

    What am I doing wrong?
    My Procedure:
    P(x) = -1 Q(x) = 4e^x

    I = e^-x

    Integral both sides = > e^-x(dy/dx -y) = e^(-x) (4e^x)dx

    = e^-x y = 4x + c

    = y = (4x + c )e^x
     
  7. Oct 16, 2009 #6

    Dick

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    You didn't use the boundary condition y(0)=4 to find C.
     
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