Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: DIFFEQ - Method of Undetermined Coefficients

  1. Sep 25, 2005 #1
    Sup' all?
    Ok, I have a quick question (hopefully). I'm trying to use the method of undetermined coefficients, and I keep getting stuck at one specific spot in the method. I'm not exactly sure what I'm doing. Let me try and explain:

    The problem is given as:
    [tex]2y''+3y'+y=t^2+3*\sin t[/tex]

    Which leads to:
    [tex]y=y_p+y_c|y_c=c_1 e^{\frac{-t}{2}} + c_2 e^{-t}[/tex]

    Now, I'm sure that the [tex]y_c[/tex] portion is correct. It is the [tex]y_p[/tex] part that I get confused on.
    I'll go through my steps, so you can see what I am doing right/wrong.

    So, we first split [tex]y_p[/tex] as follows:

    [tex]y_p=y_{p1}+y_{p2} [/tex]
    Where:
    (*1) - [tex] y_{p1} [/tex] satisfies [tex]2y_{p1}''+3y_{p1}'+y_{p1}=t^2[/tex]
    (*2) - [tex] y_{p2} [/tex] satisfies [tex]2y_{p2}''+3y_{p2}'+y_{p2}=3\sin t [/tex]

    For the [tex] y_{p1} [/tex] portion:

    [tex] y_{p1} = At^2+Bt+C [/tex]
    [tex] y_{p1}'' = 2At+B [/tex]
    [tex] y_{p1}'' = 2A [/tex]

    Plugging into (*1) yields:

    [tex] 2[2A]+3[2At+B]+[At^2+Bt+C] = t^2 [/tex]
    [tex] [A]t^2 + [6A+B]t^1 +[4A+3B+C]t^0 = t^2 [/tex]

    Now this is where I get confused.
    I'm supposed to factor and arrange the terms, and setup a system of equations?

    So, maybe something like this?

    [tex] t^2: A = \lambda_1 [/tex]
    [tex] t^1: 6A + B = \lambda_2 [/tex]
    [tex] t^0: 4A+3B+C= \lambda_3 [/tex]

    Now, how do I know what [tex]\lambda_n [/tex] are? The book, seems to magically find a number for them, but I'm NOT sure where those numbers are coming from. So if someone could explain this step, I would be very thankful. I think once I understand this step that I will be able to carry on with the other problems and do the [tex] y_{p2} [/tex] portion of this problem also.

    Thanks in advance :)
     
  2. jcsd
  3. Sep 25, 2005 #2

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Do it this way:

    So you're left with:

    [tex]2(2A)+3(2At+B)+At^2+Bt+C=t^2[/tex]

    expand it out:

    [tex]4A+6At+3B+At^2+Bt+C=t^2[/tex]

    Now combine like factors:

    [tex](4A+3B+C)+(6A+B)t+At^2=t^2[/tex]

    Now equate the coefficients on both side for like terms:

    A=1
    6A+B=0
    4A+3B+C=0

    So solve for A, B, and C and do the same for the other one.
     
    Last edited: Sep 25, 2005
  4. Sep 25, 2005 #3
    I don't understand what "equate the coefficients on both sides for like terms" means. I'm probably missing something really basic here. But, how are you getting the {1,0,0} in the system of equations?
     
  5. Sep 25, 2005 #4

    lurflurf

    User Avatar
    Homework Helper

    The lambda are the coefficents. Equate them to the given coeffficients and solve for the unknowns using algebra.
     
  6. Sep 25, 2005 #5
    No offense lurflurf, but I just said that I don't understand what "equate the coefficients means. So telling me to equate the coefficients again doesn't make any sense to me.

    I'm just guessing here, but maybe it means this ?

    Ok, so our factored equation with the appropriate [tex]y_{pn}^{(m)}[/tex] plugged in is:
    [tex](4A+3B+C)+(6A+B)t+At^2=t^2[/tex]

    So equating the coefficients means:
    [tex] t^n : P(N) = \lambda_n [/tex]
    Where [tex] \lambda_n [/tex] means the count of coefficients in the expression we are trying to find a function for.

    So the function we are trying to find a [tex] y_{pn} [/tex] function for is [tex] t^2 [/tex], which can be written as:

    [tex] c_1t^2 + c_2t^1 + c_3t^0 |c_1=1,\,c_2=0,\,c_3=0[/tex]

    So that is how you get the [tex] \lambda_n [/tex] expressions. So, equating the coefficients really does make sense after all.

    Sorry lurflurf!!!

    hehe, how could I have been so stupid :)

    that's it? Right? grr.... I hope that's what it is, I'll have to try it in a second with a different problem.
     
  7. Sep 26, 2005 #6

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    I'm probably too late but for the record here goes: So we have:

    [tex](4A+3B+C)+(6A+B)t+At^2=t^2[/tex]

    That's the same thing as :

    [tex](4A+3B+C)t^0+(6A+B)t+At^2=0t^0+0t+1t^2[/tex]

    Now, for this to be an equality for EVERY value of t, the coefficients on each power of t must be the same. So look at the coefficient on [itex]t^0[/itex]

    That means:
    4A+3B+C=0

    right?

    Same dif for t ok?
    6A+B=0

    Same for t^2:

    A=1.

    What about:

    [tex](2a-4b+6a-12)+(22a-c)x+(-c+3d)x^2-(16c-22d)x^3=1-2x^2-x^3[/tex]

    Can you set up the the four equations which "equate" coefficients? (Don't solve them as this is just a made-up problem ok)
     
  8. Sep 26, 2005 #7
    [tex] 2a-4b+6a-12 = 1[/tex]
    [tex] 22a-c = 0[/tex]
    [tex] -c+3d = -2[/tex]
    [tex] 16c-22d = -1[/tex]

    :)

    Thanks saltydog, I'm pretty sure I understand it. I seriously appreciate the help. I just didn't understand how the book was "magically" getting those numbers in the system of equations. So, thanks guys.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook