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Math Help - Differential equation xy'+y= sinx

  1. #1
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    Differential equation xy'+y= sinx

    xy'+y= sinx IVP is y(pi)=0

    My attempt:

    y' +y/x = sinx/x

    Finding the I.F

    e^$1/x * dx = e^lnx

    I.F is x


    Muliply all the equation by x

    xy' + y = sinx

    Restate the equation

    (x*y)' = sinx

    xy = -cosx + c

    y = -cosx / x + c

    I think i have done right so far, but I dont know how to to evaluate it with the
    values look:

    0 = - cos pi / pi + c ??? what How do i evaluate that?
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  2. #2
    Super Member TheChaz's Avatar
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    -cos(pi)/pi = 1/pi
    So your equation becomes
    0 = 1/pi + c
    c = -1/pi
    y = -cos(x)/x - 1/pi

    (edit)
    The above corresponds to a previous error in calculation, but I have left it so the following posts don't appear unmotivated
    Last edited by TheChaz; March 12th 2011 at 09:41 AM. Reason: ibid
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  3. #3
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    It's actually \displaystyle y = -\frac{\cos{x}}{x} + \frac{c}{x}...
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    what happened to 1/pi ? how did it become c/ x? didnt get that
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by Riazy View Post
    what happened to 1/pi ? how did it become c/ x? didnt get that
    Prove it just went back to the point right after you integrated

    xy=-\cos(x)+C Now if you isolate y you get

    \displaystyle y=-\frac{\cos(x)}{x}+\frac{C}{x} Now we can use the initial conditions to solve for C

    \displaystyle 0=-\frac{\cos(\pi)}{\pi}+\frac{C}{\pi} \implies C=-1
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