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Math Help - DIFFERENTIAL EQUATIONS -Can you help me ?

  1. #1
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    Angry DIFFERENTIAL EQUATIONS -Can you help me ?

    Verify that y = sinxcosx - cosx is a solution of the initial-value problem y' + (tanx)y= cos^2x , y(0) = -1 on the interval - - П/2 < x< П/2
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  2. #2
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    Quote Originally Posted by butbi9x View Post
    Verify that y = sinxcosx - cosx is a solution of the initial-value problem y' + (tanx)y= cos^2x , y(0) = -1 on the interval - - П/2 < x< П/2
    Hi. I'm new to differential equations; however, I think I see the solution. Check it out:

    The idea here is to try and make one side look like the derivative of a product. Take the left side and multiply by the function sec X. Carrying that over the entire equation yields: SecX*y' + secXtanX*y =cosX

    This is clearly the deritive of the product secX*y. Simply integrate both sides and get the new equation: secX*y=sinX +c. Just divide by secX and you receive: Y=sinXcosX+cosX
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  3. #3
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    Use method of linear equations

    find integrating factor first: IF=exp(integral(p(x)))

    exp(integral(tanx))=secx

    y=cosx(integral(cosx))

    y=sinxcosx+kcosx

    use initial value: -1=k

    y=sinxcosx-cosx or...

    y=(1/2)sin2x -cosx
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