Results 1 to 4 of 4

Math Help - [SOLVED] non-homogeneous differential eqn. - method of undeterminded coefficients

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    82

    [SOLVED] non-homogeneous differential eqn. - method of undeterminded coefficients

    Here is the equation:
    y''+3y'+2y=2cos(2x)-6sin(2x)

    I am supposed to solve this differential equation using 2 different methods. Method 1, I understand fine, but method 2 is what I need help understanding.

    METHOD 1: Find complete solution using real trial functions. By using Asin(2x)+Bcos(2x), differentiating it and putting it back into the origional equation I got the answer of y=C_{1} e^{-x}+C_{2} e^{-2}+\frac{3}{5}sin(2x)+\frac{4}{5}cos(2x) which IS correct. I got this method down.

    METHOD 2: Here are my given directions for "using a complex valued approach"
    a) Solve the equation y''+3y'+2y=2cos(2x)-6sin(2x)=e^{2ix} for a patricular solution z_{p}=c e^{2ix}, where c is a complex constant.
    b) Expanding z_{p}=c e^{2ix}=z_{1}(x)+i z_{2}(x);
    c)Noting that: z_{1}(x)=Re[ce^{2ix}] and z_{2}(x)=Im[ce^{2ix}] are solutions to both solutions to cos(2x) and sin(2x) respectively.
    d)Now by taking the right combination of z_{1}(x), z_{2}(x) to obtain a particular solution to the original differential equation.

    Hopefully I made what I need to do in method 2 clear, unfortuneatly, I am not clear on how to do it. If anyone would be willing to run throught the solution procedure using this "complex valued approach" I would be thankful.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Start with the equation y''+3y'+2y=e^{2ix} and we know a particular solution to the homogeneous equation y''+3y+2y=0 is y_1=e^{-x}. Now let y=e^{-x}v and w=v' and run through that routine in the book to get down to the expression e^{-x}w'+e^{-x}w=e^{x(2+2i)}. Integrate it once to get w=\frac{1}{2i+2}e^{x(2i+1)}+c_1 e^{-x}. Integrate it again to get v=-c_1 e^{-x}+\frac{1}{(2i+2)(2i+1)}e^{x(2i+1)}+c2. Back-substitute to obtain the solution

    y=c_1 e^{-x}-c_2 e^{-2x}+\frac{1}{(2i+2)(2i+1)}e^{2ix}

    to the non-homogeneous equation.

    Now the part with c1 and c2 represent the solution to the homogeneous case so that a particular solution is just the other part which I can write is as:

    y_p(x)=\frac{1}{(2i+2)(2i+1)}\left(\cos(2x)+i\sin(  2x)\right)

    Now, if I expand y_p(x) into it's real and imaginary parts, then the real part, when substituted into the ODE, will give \cos(2x) and the imaginary part will give \sin(2x) and therefore, the function y(x)=2\text{Re}\,y_p(x)-6\text{Im}\,y_p(x) is a solution to:

    y''+3y'+2y=2\cos(2x)-6\sin(2x)
    Last edited by shawsend; March 15th 2010 at 05:11 PM. Reason: Corrected DE in w
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2009
    Posts
    82
    Thanks for the help. I just need a little clarification.

    Quote Originally Posted by shawsend View Post
    Now let y=e^{-x}v and w=v' and run through that routine in the book to get down to the expression e^{-x}w'+e^{-x}w=e^{x(2+2i)}.
    I understand almost everything put the part quoted above. I am unsure of how you got: e^{-x}w'+e^{-x}w=e^{x(2+2i)}. Unfortuneately, I do not have a book because my professor thinks that his handouts in class are better than a textbook when his handouts do not have a single example of this method (or any examples ) or any more explanation that what I gave you in the problem...

    We can choose any of the homogeneous solutions: e^{-x}, e^{-2x} right?
    These are not particular solutions (because they only satisfy the homogenoues equation) or not?
    Last edited by snaes; March 15th 2010 at 06:48 PM. Reason: Wording error...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Quote Originally Posted by snaes View Post
    Thanks for the help. I just need a little clarification.



    I understand almost everything put the part quoted above. I am unsure of how you got: e^{-x}w'+e^{-x}w=e^{x(2+2i)}. Unfortuneately, I do not have a book because my professor thinks that his handouts in class are better than a textbook when his handouts do not have a single example of this method (or any examples ) or any more explanation that what I gave you in the problem...

    We can choose any of the homogeneous solutions: e^{-x}, e^{-2x} right?
    These are not particular solutions (because they only satisfy the homogenoues equation) or not?
    Just substitute ye^{-x} into the original DE to arrive at the DE in w. Check out any DE text book on the subject of "Reduction of Order" and work through the example to better understand it. You can use any of the two general solutions and e^{-x} is a particular solution while c_1 e^{-x} is a general solution.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Non-homogeneous 2nd order DE with constant coefficients.
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: April 8th 2010, 01:15 PM
  2. non-homogeneous equation, method of undetermined coefficients
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: March 4th 2010, 12:53 PM
  3. Replies: 1
    Last Post: May 4th 2009, 08:21 PM
  4. Replies: 8
    Last Post: March 14th 2009, 07:29 PM
  5. second order differential constant coefficients homogeneous
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: October 1st 2008, 10:50 AM

Search Tags


/mathhelpforum @mathhelpforum