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Math Help - differential equation w/ initial condition

  1. #1
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    differential equation w/ initial condition

    Solve the differential equation with given conditions

    (dy/dt)=(1/2)y

    y(2)=100
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  2. #2
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    Quote Originally Posted by thedoge View Post
    Solve the differential equation with given conditions

    (dy/dt)=(1/2)y

    y(2)=100
    y'=\frac{1}{2}y
    Divide by y.
    First check if y=0 is a solution (it is).
    Now look for other solutions y\not = 0,
    \frac{y'}{y}=\frac{1}{2}
    Integrate,
    \int \frac{y'}{y} dx = \int\frac{1}{2} dx
    Thus,
    \ln |y|=\frac{1}{2}x+C'
    Thus, e^{\ln |y|}=e^{\frac{1}{2}x+C'}
    |y|=Ce^{\frac{1}{2}x},C>0
    y=Ce^{\frac{1}{2}x}
    100=Ce^{\frac{1}{2}(2)}
    100=Ce^{1}=Ce
    C=100/e=100e^{-1}.
    Thus,
    y=Ce^{\frac{1}{2}x}=100e^{-1}e^{\frac{1}{2}x}=100e^{\frac{1}{2}x-1}
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  3. #3
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    As fast and amazing as usual PH.

    If you don't mind me asking, do you have a specific profession outside this forum? You seem to be a master of mathematics which should open up about any occupation to you
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  4. #4
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    Quote Originally Posted by thedoge View Post

    If you don't mind me asking, do you have a specific profession outside this forum?
    Yes! I am a fashion designer.
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  5. #5
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    Haha. A comedian too.

    Unless you're serious. One can never know online;]
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  6. #6
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    Hrm. I follow everything you said perfectly and even worked it out that way myself, but apparently this problem needs to evaluate to a number.

    What is the value of x?

    Nevermind. I see what the problem was.

    *** for future reader's reference replace the x in 100*e^(.5x-1) with a t
    Last edited by thedoge; February 4th 2007 at 08:11 PM.
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