Results 1 to 3 of 3

Math Help - Solving for A,B w/ Method of Undetermined Coefficients

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    20

    Solving for A,B w/ Method of Undetermined Coefficients

    In trying to find the particular solution to a non-homogeneous third-order ODE, I used the method of undetermined coefficients but it seems to be the method of undeterminable coefficients.

    I have

    y^(3) - y^(1) = 2sint

    So I assumed

    Y(t) = Acost + Bsint

    Differentiating and plugging in, I get

    (Asint - Bcost)^3 + Asint - Bcost = 2sint

    In solving these problems before, I would follow the idea, "There is a 0cost on the right, so the collected coefficients of the cost on the left must be zero." Applying this to the (sint)^3, (sint)^2(cost), (sint)(cost)^2, (cost)^3, sint, and cost, I would think that

    A^3 = 0
    -(A^2)B = 0
    A(B^2) = 0
    -B^3 = 0
    A = 2
    -B = 0

    However, A cannot both be two and its cube be zero. Also, the back of the book gives Y(t) as cost, for which, if Y(t) = Acost + Bsint, A would be one.

    I know I am solving for the coefficients incorrectly, but I don't know why my method is wrong here (since I know it has worked in other situations). How can I solve for these coefficients? Is my assumption of what Y(t) is wrong? Should it be something other than Acost + Bsint?

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

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Oijl View Post
    In trying to find the particular solution to a non-homogeneous third-order ODE, I used the method of undetermined coefficients but it seems to be the method of undeterminable coefficients.

    I have

    y^(3) - y^(1) = 2sint

    So I assumed

    Y(t) = Acost + Bsint

    Differentiating and plugging in, I get

    (Asint - Bcost)^3 + Asint - Bcost = 2sint

    In solving these problems before, I would follow the idea, "There is a 0cost on the right, so the collected coefficients of the cost on the left must be zero." Applying this to the (sint)^3, (sint)^2(cost), (sint)(cost)^2, (cost)^3, sint, and cost, I would think that

    A^3 = 0
    -(A^2)B = 0
    A(B^2) = 0
    -B^3 = 0
    A = 2
    -B = 0

    However, A cannot both be two and its cube be zero. Also, the back of the book gives Y(t) as cost, for which, if Y(t) = Acost + Bsint, A would be one.

    I know I am solving for the coefficients incorrectly, but I don't know why my method is wrong here (since I know it has worked in other situations). How can I solve for these coefficients? Is my assumption of what Y(t) is wrong? Should it be something other than Acost + Bsint?

    Thank you!
    I guess I 'm a little confused over your notation...

    The way I read your problem is this

    \frac{d^3y}{dt^3}-\frac{dy}{dt}=2\sin(t)

    Is this is the case your auxillary equation is

    m^3-m=0 \iff m(m-1)(m+1)

    so y_c=c_1+c_2e^{t}+c_3e^{-t}

    Now y_p=A\sin(t)+B\cos(t)\

    \frac{dy_p}{dt}=A\cos(t)-B\sin(t)

    \frac{d^3y_p}{dt^3}=-A\cos(t)+B\sin(t)

    Plugging into the original we get

    -A\cos(t)+B\sin(t)-(A\cos(t)-B\sin(t))=2\sin(t)

    -2A\cos(t)+2B\sin(t)=2\sin(t)

    This tells us that A=0 and B=1

    So y_p=\cos(t)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    20
    Doh! Of course, thanks, just needed another pair of eyes. I had plugged the derivatives of the particular solution into the auxiliary equation, somehow. I need better bookkeeping while doing problems, I guess. Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Method of undetermined coefficients
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: November 29th 2011, 03:59 PM
  2. Method of Undetermined coefficients
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: October 23rd 2009, 06:05 PM
  3. Method of Undetermined Coefficients
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: March 10th 2009, 07:34 PM
  4. Method of Undetermined Coefficients
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 14th 2008, 02:15 AM
  5. Method of Undetermined Coefficients
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 15th 2008, 02:48 PM

Search Tags


/mathhelpforum @mathhelpforum