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Thread: Weird 2nd Order difeq question. Help Please!!

  1. #1
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    Weird 2nd Order difeq question. Help Please!!

    Here is the question:

    y'' = (y')^2 (Assume y and y' are positive)

    I looked at the question and didn't know how to start it, never seen a question in this form. Help please!!!
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  2. #2
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    Quote Originally Posted by Phyxius117 View Post
    Here is the question:

    y'' = (y')^2 (Assume y and y' are positive)

    I looked at the question and didn't know how to start it, never seen a question in this form. Help please!!!
    Divide by $\displaystyle y'$ and integrate once

    $\displaystyle \frac{y''}{y'} = y' $ so $\displaystyle \ln y' = y + \ln c_1 \; \text{or}\; y' = c_1 e^y.$

    Then separate and integrate again.
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  3. #3
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    Thanks!!

    Now I got

    y = C1 e^(y) X + C2

    as answer but how do I separate to just y in the equation with e^(y) present?
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  4. #4
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    lol I think it's wrong but I plug

    y = lny' - C1 into y

    and becomes

    y= C1 e^(lny' - C1) + C2

    which simplifies into

    y = C1y' + C2

    then I integrate again becomes

    yx = C1y + C2x + C3

    and finally

    y = (C2x + C3) / (X - C1)

    which I feel it's wrong...
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  5. #5
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    Quote Originally Posted by Phyxius117 View Post
    lol I think it's wrong but I plug

    y = lny' - C1 into y

    and becomes

    y= C1 e^(lny' - C1) + C2

    which simplifies into

    y = C1y' + C2

    then I integrate again becomes

    yx = C1y + C2x + C3

    and finally

    y = (C2x + C3) / (X - C1)

    which I feel it's wrong...
    Ya - you can't do that!

    Quote Originally Posted by Danny View Post
    Divide by $\displaystyle y'$ and integrate once

    $\displaystyle \frac{y''}{y'} = y' $ so $\displaystyle \ln y' = y + \ln c_1 \; \text{or}\; y' = c_1 e^y.$

    Then separate and integrate again.
    From what I have.

    $\displaystyle e^{-y} dy = c_1 dx$ so $\displaystyle -e^{-y} = c_1 x + c_2$ (I absorbed the negative into the constants).

    Solving for y gives y$\displaystyle = - \ln | c_1 x + c_2|$.
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  6. #6
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    Just put
    y'=f.

    The equation then becomes
    $\displaystyle
    \frac{df}{dx}=f^2(x)
    $

    Separate and integrate to get

    $\displaystyle
    f^{-2}df=dx
    $
    and
    $\displaystyle
    -f^{-1}=x+a
    $,
    Hence
    $\displaystyle
    f(x)=-\frac{1}{x+a}.
    $

    Now,
    $\displaystyle
    y'=-\frac{1}{x+a}
    $

    Integrate to get the final answer
    $\displaystyle
    y(x)=-\ln(|x+a|)+b
    $
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  7. #7
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    ty so much!!!
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  8. #8
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    Note that Danny's answer is also correct:

    y
    $\displaystyle
    y=-\ln|c_1x+c_2|=-\ln(|c_1(x+\frac {c_2}{c_1}|)=-\ln(|c_1|)-\ln(|x+\frac{c_2}{c_1}|)=-\ln(|x+a|)+b
    $
    where b=-ln(c1) and a=c2/c1
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