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Math Help - First Order Derivative II

  1. #1
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    First Order Derivative II

    \frac{\mathrm{d}y}{\mathrm{d}x} = y - 2, \ \ \ y(0)=1

    Is this solveable to get y(x)=.... When I used seperation of variable I got c=\ln (-1) which isn't possible. Can someone solve it? Thanks in advance.
    Last edited by Simplicity; February 18th 2009 at 01:03 PM. Reason: Wrong Value
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  2. #2
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    Hi

    Don't forget the absolute value !

    Integrating \frac{u'(x)}{u(x)} gives \ln(|u(x)|)
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    Don't forget the absolute value !

    Integrating \frac{u'(x)}{u(x)} gives \ln(|u(x)|)
    Is this even case when:

    -1 = e^c

    \ln (-1) = c

    It's not from the integration so it is still \ln (-1) = \ln (1)?
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  4. #4
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    No, look :

    \frac{dy}{dx} = y - 2, \ \ \ y(0)=-1

    \frac{dy}{y-2} = dx, \ \ \ y(0)=-1

    \ln(|y(x)-2|) = x + c, \ \ \ y(0)=-1

    \ln(|y(0)-2|) = c, \ \ \ y(0)=-1

    \ln(3) = c
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  5. #5
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    Quote Originally Posted by running-gag View Post
    No, look :

    \frac{dy}{dx} = y - 2, \ \ \ y(0)=-1

    \frac{dy}{y-2} = dx, \ \ \ y(0)=-1

    \ln(|y(x)-2|) = x + c, \ \ \ y(0)=-1

    \ln(|y(0)-2|) = c, \ \ \ y(0)=-1

    \ln(3) = c
    No, Sorry, I wrote the question wrong:

    \frac{dy}{dx} = y - 2, \ \ \ y(0)=1

    It's 1 and not -1.
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  6. #6
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    And \ln 1 = 0.

    Which would give equation y=x+2 which when you differentiate does not give the required solution.

    \frac{\mathrm{d}y}{\mathrm{d}x} = y - 2, \ \ \ y(0)=1

    Can someone please solve my confusion.
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  7. #7
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    \frac{dy}{dx} = y - 2, \ \ \ y(0)=1

    \frac{dy}{y-2} = dx, \ \ \ y(0)=1

    \ln(|y(x)-2|) = x + c, \ \ \ y(0)=1

    \ln(|y(0)-2|) = c, \ \ \ y(0)=1

    \ln(1) = c

    c = 0

    \ln(|y(x)-2|) = x, \ \ \ y(0)=1

    |y(x)-2| = e^x, \ \ \ y(0)=1

    y(x)-2 = e^x or -y(x)+2 = e^x

    y(x) = e^x + 2 or y(x) = -e^x + 2

    y(0) = 1 therefore y(x) = -e^x + 2
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