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Math Help - Calculus quiz tomorrow..help with this one review problem?

  1. #1
    Newbie woohoo's Avatar
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    Question Calculus quiz tomorrow..help with this one review problem?

    Solve initial value problem using sep. of variables

    y(prime)=cos^2(x) y(0)=pi/4

    I ended up using integration by parts and got to here:

    y=cosx * sinx - x + C

    and C= pi/4

    but the book says the answer is:

    y=.5(x+.5sin(2x))+pi/4

    what did I do wrong?
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  2. #2
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    Quote Originally Posted by woohoo View Post
    Solve initial value problem using sep. of variables

    y(prime)=cos^2(x) y(0)=pi/4

    I ended up using integration by parts and got to here:

    y=cosx * sinx - x + C

    and C= pi/4

    but the book says the answer is:

    y=.5(x+.5sin(2x))+pi/4

    what did I do wrong?
    Use the identity \cos^2{x} = \frac{1}{2}(1 + \cos{2x}).

    So y = \int{\cos^2{x}\,dx} = \frac{1}{2}\int{(1 + \cos{2x})\, dx}

     = \frac{1}{2}(x + \frac{1}{2}\sin{x}) + C

    Since y(0) = \frac{\pi}{4},

    \frac{\pi}{4} = \frac{1}{2}(0 + \frac{1}{2}\sin{0}) + C

    C = \frac{\pi}{4}

    So y = \frac{1}{2}(x + \frac{1}{2}\sin{x}) + \frac{\pi}{4}.
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  3. #3
    Newbie woohoo's Avatar
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    Okay I understand the book's answer now.

    But I don't get why my method turned out incorrect.

    I got to where y= integral (cos^2(x) dx)=integral (cosx * cosx dx)

    using separation by parts:

    y=cosxsinx-integral(x)=cosxsinx-x+C

    where did I go wrong?
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  4. #4
    Newbie woohoo's Avatar
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    Nevermind.

    I figure out where I went wrong.

    Thanks!
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  5. #5
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    Quote Originally Posted by woohoo View Post
    Okay I understand the book's answer now.

    But I don't get why my method turned out incorrect.

    I got to where y= integral (cos^2(x) dx)=integral (cosx * cosx dx)

    using separation by parts:

    y=cosxsinx-integral(x)=cosxsinx-x+C

    where did I go wrong?
    Are you trying to do separation of variables or integration by parts? They're two different methods.

    You only use separation of variables if you have a Separable linear DE, which you don't, so I guess you're trying to do integration by parts...

    Integration by parts is as follows...

    \int{u\,dv} = uv - \int{v\, du}.

    So if you're integrating \cos{x}\times\cos{x}, then u = \cos{x} and dv = \cos{x}. So what are du and v?

    In this case you should end up with \int{\sin^2{x}\,dx}... not fun to work with as it's just as bad as cos^2{x}. Just use the trig identity.
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