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Math Help - First order Differential Equations

  1. #1
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    First order Differential Equations

    Hi
    I am having trouble with the following questions:
    Solve the differential equations
    1)  \frac{dy}{dx} = e^{x+y}

    \frac{dy}{dx} = e^x * e^y

    \frac{1}{e^y} * \frac{dy}{dx} = e^x

    not sure what i should do next

    2) \frac{dy}{dx} = 1+x-y-xy

    y * \frac{dy}{dx}

    y(1+x) * \frac{dy}{dx} = 1+x

    y * \frac{dy}{dx} = \frac{1+x}{1+x}

    y * \frac{dy}{dx} = 1

    \frac{y^2}{2} = x+ C

    y^2 = \frac{x+C}{2}

    P.S
    Last edited by mr fantastic; August 27th 2010 at 03:36 AM. Reason: Fixed latex, re-titled.
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  2. #2
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    \displaystyle\frac{dy}{dx} = e^x \times  e^y

    \displaystyle\frac{dy}{dx}e^{-y} = e^x

    \displaystyle e^{-y}~dy = e^x~dx

    \displaystyle \int e^{-y}~dy = \int e^x~dx
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  3. #3
    MHF Contributor matheagle's Avatar
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    I'm not sure what you did with number 2, but it's linear and separable.

    {dy\over dx}=1+x-y(1+x)=(1+x)(1-y)
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  4. #4
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    Quote Originally Posted by matheagle View Post
    I'm not sure what you did with number 2, but it's linear and separable.

    {dy\over dx}=1+x-y(1+x)=(1+x)(1-y)
    where does the +x go??
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  5. #5
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    Matheagle has taken out  (1+x) as a factor. Its called factoring by grouping.

    {dy\over dx}=1+x-y(1+x)=(1+x)-y(1+x)= (1+x)(1-y)

    {dy\over dx}= (1+x)(1-y)

    {dy\over (1-y)}= (1+x)~dx

    \int {dy\over (1-y)}= \int(1+x)~dx
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  6. #6
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    This is my answers but it doesn't match the answer's in book

    1)
    y-e^{-y} = e^x+C
    e^{-y} = -(e^x+C)
    y = -ln(-e^x-C)

    2)
    ln|1-y| = x+\frac{x^2}{2} + C
    1-y = e^{x+\frac{x^2}{2} + C}
    -y = -1+e^{x+\frac{x^2}{2} + C}
    y=1-e^{x+\frac{x^2}{2} + C}
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  7. #7
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    Quote Originally Posted by Paymemoney View Post
    This is my answers but it doesn't match the answer's in book

    1)
    y-e^{-y} = e^x+C
    e^{-y} = -(e^x+C)
    y = -ln(-e^x-C)

    2)
    ln|1-y| = x+\frac{x^2}{2} + C
    1-y = e^{x+\frac{x^2}{2} + C}
    -y = -1+e^{x+\frac{x^2}{2} + C}
    y=1-e^{x+\frac{x^2}{2} + C}
    What are your book's answers?

    Note:

    -C can be 're-branded' as B,
    e^C can be 're-branded' as A

    since in both cases C is arbitrary and therefore B and A are also arbitrary.
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  8. #8
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    Book's Answers are
    1) actually got it right y=-ln(k-e^x) where k=-c

    2) y=1+ke^{-(x+\frac{x^2}{2})}
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  9. #9
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    Quote Originally Posted by Paymemoney View Post
    [snip]
    2)
    ln|1-y| = x+\frac{x^2}{2} + C Mr F says: There should be a negative in front of the ln. Make the necessary corrections in your working below and then recall from my previous post the note about 're-branding' C ....

    1-y = e^{x+\frac{x^2}{2} + C}
    -y = -1+e^{x+\frac{x^2}{2} + C}
    y=1-e^{x+\frac{x^2}{2} + C}
    ..
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  10. #10
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    i don't understand why should there be a negative at the front of ln?

    The intergal of \frac{1}{1-y} = ln|1-y|
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  11. #11
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    \displaystyle{\int\frac{dy}{1-y}=-\int\frac{dy}{y-1}=-\ln|y-1|+C.}

    The rule you're thinking of is that

    \displaystyle{\int\frac{dy}{y-a}=\ln|y-a|+C.}
    However, the sign of y must be positive for that rule to work.
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  12. #12
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    Quote Originally Posted by Ackbeet View Post
    \displaystyle{\int\frac{dy}{1-y}=-\int\frac{dy}{y-1}=-\ln|y-1|+C.}

    The rule you're thinking of is that

    \displaystyle{\int\frac{dy}{y-a}=\ln|y-a|+C.}
    However, the sign of y must be positive for that rule to work.
    oh i see now
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