Results 1 to 5 of 5

Math Help - 2nd ODE Homogeneous, Double roots

  1. #1
    Member
    Joined
    Apr 2010
    Posts
    81

    2nd ODE Homogeneous, Double roots

    y^{\prime \prime} + 6 y^{\prime} + 9 y = 0

    Initial Conditions:
    y(0)=3,\quad y^{\prime}(0)=-7

    So, I determined the general solution should look like this:

    Ae^{-3x}+Be^{-3x}=0

    When Y(0)=3
    [Math]3=A+B[/tex]

    And when Y'(0)=-7
    <br />
-7=-3A-3B<br />

    Im stuck here. Anytime I try and solve for one variable, I get
    -7=-9 which is a totally false statement.

    What do I do from here?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,656
    Thanks
    1480
    When there's a repeated root, the solution is actually \displaystyle Ae^{-3x} + Bx\,e^{-3x}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2010
    Posts
    81
    Quote Originally Posted by Prove It View Post
    When there's a repeated root, the solution is actually \displaystyle Ae^{-3x} + Bx\,e^{-3x}.
    So when Y(0)=3
    3=A

    and Y'=-3Ae^{3x}+[Be^{-3x}-3Be^{-3x}]
    So when Y'(0)=-7,
    B=3

    So my final solution is
    3e^{-3x}+3xe^{-3x}

    This seems to be wrong! Where did I mess up?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2
    Let
    y(x)=(a+bx)\mathrm{e}^{-3x},
    then
    y'(x)=((b-3a)-3bx)\mathrm{e}^{-3x}.
    Applying the initial conditions, we get
    a=3 and b-3a=-7,
    which gives you
    a=3 and b=2.
    Therefore, your solution is
    y(x)=3\mathrm{e}^{-3x}+2x\mathrm{e}^{-3x}.

    You seem to compute b wrong.
    Last edited by bkarpuz; March 28th 2011 at 09:09 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,656
    Thanks
    1480
    The derivative is actually

    \displaystyle y'(x) = -3Ae^{-3x} + Be^{-3x} - 3Bx\,e^{-3x}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 15th 2010, 05:43 AM
  2. Non-Homogeneous PDE
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: November 6th 2010, 06:19 AM
  3. Homogeneous de
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 4th 2008, 05:08 AM
  4. Non-Homogeneous DQ's
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: July 8th 2008, 05:10 PM
  5. Given 2 imaginary roots find other 2 roots
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 26th 2008, 09:24 PM

Search Tags


/mathhelpforum @mathhelpforum