Results 1 to 8 of 8

Math Help - Weird 2nd Order difeq question. Help Please!!

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    65

    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!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,325
    Thanks
    10
    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 y' and integrate once

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

    Then separate and integrate again.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2008
    Posts
    65
    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?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2008
    Posts
    65
    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...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,325
    Thanks
    10
    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 y' and integrate once

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

    Then separate and integrate again.
    From what I have.

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

    Solving for y gives y  = - \ln | c_1 x + c_2|.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Nov 2009
    Posts
    13
    Just put
    y'=f.

    The equation then becomes
    <br />
\frac{df}{dx}=f^2(x)<br />

    Separate and integrate to get

    <br />
f^{-2}df=dx<br />
    and
    <br />
-f^{-1}=x+a<br />
,
    Hence
    <br />
f(x)=-\frac{1}{x+a}.<br />

    Now,
    <br />
y'=-\frac{1}{x+a}<br />

    Integrate to get the final answer
    <br />
y(x)=-\ln(|x+a|)+b<br />
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Nov 2008
    Posts
    65
    ty so much!!!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Nov 2009
    Posts
    13
    Note that Danny's answer is also correct:

    y
    <br />
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<br />
    where b=-ln(c1) and a=c2/c1
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. weird question
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: May 24th 2010, 09:30 PM
  2. weird question
    Posted in the Statistics Forum
    Replies: 3
    Last Post: March 13th 2010, 12:03 PM
  3. weird question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 29th 2009, 09:47 AM
  4. Quasilinear first order PDE-weird parametrization?!
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: September 28th 2009, 01:28 PM
  5. check my work please (difEq)
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: January 31st 2009, 01:46 PM

Search Tags


/mathhelpforum @mathhelpforum