Results 1 to 7 of 7

Math Help - Another differential equation

  1. #1
    Member
    Joined
    Jul 2008
    Posts
    119

    Another differential equation

    Explicitly find the maximal interval I about 0 on which we can solve the following differential equation:


    I'm not sure how to approach. I know that if it was


    then I can divide both sides and integrate and get
    , but there's a square term.

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Paperwings View Post
    Explicitly find the maximal interval I about 0 on which we can solve the following differential equation:


    I'm not sure how to approach. I know that if it was


    then I can divide both sides and integrate and get
    , but there's a square term.

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2008
    Posts
    119
    Ah, ok. Thanks mathdude.

    1.
    2. Integrating yields

    3. Solving for f(x) gives

    However, substituting the initial condition gives

    I don't understand this part. Wouldn't this function be undefined at zero?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jul 2008
    From
    Sofia, Bulgaria
    Posts
    75
    Quote Originally Posted by Paperwings View Post
    Ah, ok. Thanks mathdude.

    1.
    2. Integrating yields

    3. Solving for f(x) gives

    However, substituting the initial condition gives

    I don't understand this part. Wouldn't this function be undefined at zero?

    Why not do the last transformation the easy way:



    \f(x) = \frac{\-1}{x+C}\\-1 = \frac{\-1}{0+C}\\C = \1
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Assume there is a non-vanishing solution on an interval I about 0.
    Thus, y' = (y)^2 this means y'/(y^2) = 1, integrating means (1/y) = x+k.
    Since y(0)=1 it means 1/y(0) = k and so k =1.
    Thus, the solution must be 1/y=x+1.
    Thus, if -1 is not in I then we can take reciprocals to get y = 1/(x+1).
    Therefore, I has the condition that -1 is not in it.
    Hence, the largest interval is (-1, \infty).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Thus, y' = (y)^2 this means y'/(y^2) = 1, integrating means (1/y) = x+k.
    Isn't it -1/y ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    It's a typo, that's all.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Partial Differential Equation satisfy corresponding equation
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: May 16th 2011, 07:15 PM
  2. Replies: 4
    Last Post: May 8th 2011, 12:27 PM
  3. Replies: 1
    Last Post: April 11th 2011, 01:17 AM
  4. [SOLVED] Solve Differential equation for the original equation
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: February 21st 2011, 01:24 PM
  5. Partial differential equation-wave equation - dimensional analysis
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: August 28th 2009, 11:39 AM

Search Tags


/mathhelpforum @mathhelpforum