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Math Help - Differential problem

  1. #1
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    Differential problem

    show that
    y = -1/2x . cosx

    is/isn't a solution to y" + y = sinx
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  2. #2
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    Quote Originally Posted by puravida12 View Post
    show that
    y = -1/2x . cosx

    is/isn't a solution to y" + y = sinx
    This is unreadable.

    Is it -\frac{1}{2}x\cos{x}, or is it -\frac{1}{2x}\cos{x}, or is it -\frac{1}{2x\cos{x}}?

    Please use LaTeX or put brackets where they're needed.
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  3. #3
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    -(1/2)x . cosx

    I apologize, it has been a long night of calculus.
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  4. #4
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    If it's a solution, then

    y = -\frac{1}{2}x\cos{x}


    y' = -\frac{1}{2}(\cos{x} - x\sin{x})

     = \frac{1}{2}x\sin{x} - \frac{1}{2}\cos{x}.


    y'' = \frac{1}{2}(\sin{x} + x\cos{x}) + \frac{1}{2}\sin{x}

     = \frac{1}{2}\sin{x} + \frac{1}{2}x\cos{x} + \frac{1}{2}\sin{x}

     = \frac{1}{2}x\cos{x} + \sin{x}.



    So y'' + y = \frac{1}{2}x\cos{x} + \sin{x} - \frac{1}{2}x\cos{x}

     = \sin{x}.


    Do you think it's a solution?
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