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Dif.eq. with trigonometric functions involving y

  1. May 22, 2012 #1
    I tried to solve it but confused. Pls. help me to solve this equation:

    dy/dx + (e^x)*Sec(y) = Tan(x);

    (hint: integrating factor is e^-ax, and a is unknown, a ε ℝ, find it, solve the equation)

    Thnx.
     
  2. jcsd
  3. May 22, 2012 #2

    tiny-tim

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    welcome to pf!

    hi mausmust! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)

    show us your full calculations, and then we'll see what went wrong, and we'll know how to help! :smile:
     
  4. May 22, 2012 #3
    Hii, thank you :)

    I did:

    dy + (exSec(y)-Tan(x))dx = 0

    0 ≠ exSec(y)Tan(y) (∂N/∂x ≠ ∂M/∂y) So, we need an integrating factor. If we use e-ax;

    -ae-ax = e(1-ax)Sec(y)Tan(y) appears.

    Shouldn't be "a" is a real value? Also;

    I tried to modify equation with ArcSecant function to become a linear equation with this form;

    dy/dx + P(x)y = Q(x)

    but it went more complicated. How can we solve it?
     
  5. May 22, 2012 #4
    I don't think you're doin' that right. Also, don't use capitals letters for the trig functions. If you have:

    [tex]Mdx+Ndy=0[/tex]

    and:

    [tex]\frac{1}{N}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)=f(x)[/tex]

    then the integrating factor is:

    [tex]u=\exp\left(\int f(x)dx\right)[/tex]

    but that's not just [itex]e^{-ax}[/itex] for real a.
     
    Last edited: May 22, 2012
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