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Thread: First Order Derivative II

  1. #1
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    First Order Derivative II

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

    Is this solveable to get $\displaystyle y(x)=...$. When I used seperation of variable I got $\displaystyle c=\ln (-1)$ which isn't possible. Can someone solve it? Thanks in advance.
    Last edited by Simplicity; Feb 18th 2009 at 12:03 PM. Reason: Wrong Value
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  2. #2
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    Hi

    Don't forget the absolute value !

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

    $\displaystyle -1 = e^c$

    $\displaystyle \ln (-1) = c$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    $\displaystyle \ln(1) = c$

    $\displaystyle c = 0$

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

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

    $\displaystyle y(x)-2 = e^x$ or $\displaystyle -y(x)+2 = e^x$

    $\displaystyle y(x) = e^x + 2$ or $\displaystyle y(x) = -e^x + 2$

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