Results 1 to 5 of 5

Thread: 2nd ODE Homogeneous, Double roots

  1. #1
    Member
    Joined
    Apr 2010
    Posts
    81

    2nd ODE Homogeneous, Double roots

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

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

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

    $\displaystyle Ae^{-3x}+Be^{-3x}=0$

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

    And when Y'(0)=-7
    $\displaystyle
    -7=-3A-3B
    $

    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
    12,880
    Thanks
    1946
    When there's a repeated root, the solution is actually $\displaystyle \displaystyle Ae^{-3x} + Bx\,e^{-3x}$.
    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 $\displaystyle \displaystyle Ae^{-3x} + Bx\,e^{-3x}$.
    So when Y(0)=3
    3=A

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

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

    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
    $\displaystyle y(x)=(a+bx)\mathrm{e}^{-3x}$,
    then
    $\displaystyle y'(x)=((b-3a)-3bx)\mathrm{e}^{-3x}$.
    Applying the initial conditions, we get
    $\displaystyle a=3$ and $\displaystyle b-3a=-7$,
    which gives you
    $\displaystyle a=3$ and $\displaystyle b=2$.
    Therefore, your solution is
    $\displaystyle y(x)=3\mathrm{e}^{-3x}+2x\mathrm{e}^{-3x}$.

    You seem to compute $\displaystyle b$ wrong.
    Last edited by bkarpuz; Mar 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
    12,880
    Thanks
    1946
    The derivative is actually

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

Similar Math Help Forum Discussions

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

Search Tags


/mathhelpforum @mathhelpforum