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First ODE

  1. Sep 13, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex]y'+ycot(x)=cos(x)[/tex]


    2. Relevant equations



    3. The attempt at a solution
    First I found the integrating factor:
    [tex]\rho (x)=e^{\int cot(x)dx}=e^{ln(sinx)}=sinx[/tex]
    Plug into the equation for first order DE...
    [tex]\int \frac{d}{dx} ysinx=\int cosxsinx dx[/tex]
    End up with:
    [tex]ysinx=\frac{-cos^{2}x}{2}+C\\y=\frac{-cosxcotx}{2}+\frac{C}{sinx}[/tex]

    I think that's wrong tho..
     
  2. jcsd
  3. Sep 13, 2013 #2

    pasmith

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    Homework Helper

    You have the right answer. But since [itex]\cos^2 x + \sin^2 x = 1[/itex], it follows that [itex](\sin^2 x)' = -(\cos^2 x)'[/itex]. Perhaps it would have been better to take [itex]\sin x \cos x = \frac12(\sin^2 x)'[/itex], since you have [itex](y\sin x)'[/itex] on the other side of the equation.

    You can in any event use [itex]\cos^2 x + \sin^2 x = 1[/itex] to simplify your answer:
    [tex]
    \frac{\cos^2 x}{\sin x} = \frac{1 - \sin^2 x}{\sin x} = \frac{1}{\sin x} - \sin x[/tex]
     
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