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Math Help - Initial Value Problem

  1. #1
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    Initial Value Problem

    Verify that y=\sin(x)\cos(x)-\cos(x) is solution of the initial value problem:

    y'+y\tan(x)=\cos^2(x)

    y(0)=-1

    -\frac{\pi}{2}<x<\frac{\pi}{2}
    --------------------------------------------------------------------------------------------------
    I don't understand the object of this problem. Obviously the differential equation gives the derivative:

    y'=\cos^2(x)-y\tan(x)

    =\cos^2(x)-\left(\sin(x)\cos(x)-\cos(x)\right)\tan(x)

    =\cos^2(x)-\sin^2(x)-\sin(x)

    Then if I differentiation y=\sin(x)\cos(x)-\cos(x) I obtain \cos^2(x)-\sin^2(x)+\sin(x). This doesn't agree with the other equation, although it is close.
    Last edited by adkinsjr; December 21st 2009 at 12:10 AM.
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  2. #2
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    Quote Originally Posted by adkinsjr View Post
    Verify that y=\sin(x)\cos(x)-\cos(x) is solution of the initial value problem:

    y'+y\tan(x)=\cos^2(x)

    y(0)=-1

    -\frac{\pi}{2}<x<\frac{\pi}{2}
    --------------------------------------------------------------------------------------------------
    I don't understand the object of this problem. Obviously the differential equation gives the derivative:

    y'=\cos^2(x)-y\tan(x)

    =\cos^2(x)-\left(\sin(x)\cos(x)-\cos(x)\right)\tan(x)

    =\cos^2(x)-\sin^2(x)-\sin(x)

    Then if I differentiation y=\sin(x)\cos(x)-\cos(x) I obtain \cos^2(x)-\sin^2(x)+\sin(x). This doesn't agree with the other equation, although it is close.

    I don't understand what you did...apparently you solved the differential equation, but you didn't have to do that. You had to check that y=\cos x\sin x-\sin x is a solution to y'+y\tan x=\cos^2x , with y(0)=-1 , and it's easy to check that it is so...

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    I don't understand what you did...apparently you solved the differential equation, but you didn't have to do that. You had to check that y=\cos x\sin x-\sin x is a solution to y'+y\tan x=\cos^2x , with y(0)=-1 , and it's easy to check that it is so...

    The second term is -sin(x)
    Tonio
    I see what I did wrong. I substituted y=\cos(x)\sin(x)-\sin(x) into the differential equation and then solved for the dervivative. Then I differeniated y=\cos(x)\sin(x)-\sin(x) to verify that it's a solution. I made a simple algebraic error in here:




    This should be \cos^2(x)-\sin^2(x)+\sin(x).

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