Results 1 to 4 of 4

Math Help - x'' + 3x' + 2x = 0

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    29

    x'' + 3x' + 2x = 0

    Show that for the system x'' + 3x' + 2x = 0
    x(0) = 0 x'(0) dosen't equal 0
    it holds that x(t) dosen't equal 0 for all t.

    I've got as far as y = Ae^-2x + Be^-x
    and the differential of this.

    Now im really stuck.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,341
    Thanks
    1247
    Quote Originally Posted by ardam View Post
    Show that for the system x'' + 3x' + 2x = 0
    x(0) = 0 x'(0) dosen't equal 0
    it holds that x(t) dosen't equal 0 for all t.

    I've got as far as y = Ae^-2x + Be^-x
    and the differential of this.

    Now im really stuck.
    OK, first off, clearly x(t) DOES equal 0 at a value of t, namely t = 0. So I assume you meant that it holds that x(t) doesn't equal 0 for all OTHER t.


    You've gotten as far as

    x = Ae^{-2t} + Be^{-t}.

    Using the initial condition

    0 = A + B \implies B = -A.

    So x = Ae^{-2t} - Ae^{-t}

    x = A(e^{-2t} - e^{-t}).


    If x = 0 then either A = 0 or e^{-2t} - e^{-t} = 0. The latter only occurs at t = 0, as was already stated by the initial condition.

    So the only other way that x = 0 is if A = 0.


    Recall that \frac{dx}{dt} \neq 0 as stated originally.

    \frac{dx}{dt} = A(-2e^{-2t} + e^{-t})

    So A(-2e^{-2t} + e^{-t}) \neq 0.

    This implies that A \neq 0.


    Therefore, x \neq 0 for any t \neq 0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2008
    Posts
    29
    Quote Originally Posted by Prove It View Post
    OK, first off, clearly x(t) DOES equal 0 at a value of t, namely t = 0. So I assume you meant that it holds that x(t) doesn't equal 0 for all OTHER t.


    You've gotten as far as

    x = Ae^{-2t} + Be^{-t}.

    Using the initial condition

    0 = A + B \implies B = -A.

    So x = Ae^{-2t} - Ae^{-t}

    x = A(e^{-2t} - e^{-t}).


    If x = 0 then either A = 0 or e^{-2t} - e^{-t} = 0. The latter only occurs at t = 0, as was already stated by the initial condition.

    So the only other way that x = 0 is if A = 0.


    Recall that \frac{dx}{dt} \neq 0 as stated originally.

    \frac{dx}{dt} = A(-2e^{-2t} + e^{-t})

    So A(-2e^{-2t} + e^{-t}) \neq 0.

    This implies that A \neq 0.


    Therefore, x \neq 0 for any t \neq 0.
    Thanks now i understand what i should have done
    i have just reread the question and i should have written holds for all t>0
    does this change the question too much?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,341
    Thanks
    1247
    Quote Originally Posted by ardam View Post
    Thanks now i understand what i should have done
    i have just reread the question and i should have written holds for all t>0
    does this change the question too much?
    It doesn't change the question at all.

    I showed that x \neq 0 for all t \neq 0.

    All t > 0 is a part of all t \neq 0.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum