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Homework Help: Undetermined Coefficients, Differential Equations

  1. Jul 10, 2015 #1
    1. The problem statement, all variables and given/known data

    y'' + y =3*sin(2t) +t*cos(2t)

    Okay, so I have found the complimentary solution, and the first partial solution as listed in my work below.

    My problem is the work on the second partial solution. I have got all the derivatives plugged into the differential equation, my problem is solving for the coefficients. Ive tried setting the coefficients paired with a t and cos equal to 1, and all the rest equal to 0, yet this returns a false result. I have tried almost 20 different combinations of different coefficients equal to different things and they all return false, but I am fairly certain the differentiating was done correctly... Or was perhaps my guess of (At+B)(C*cos(2t)+D*sin(2t)) incorrect for this solution??? I don't understand what I am doing wrong.

    Main Issue: Solving for the undetermined Coefficients at the end

    2. Relevant equations

    3. The attempt at a solution

  2. jcsd
  3. Jul 10, 2015 #2
    Are you not given any initial conditions? at all? o.0
  4. Jul 10, 2015 #3


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    First off, you have shown only one solution to the characteristic equation r2 + 1 = 0. There should be two.
  5. Jul 10, 2015 #4
    The factor alongside Cosine is not constant. Shouldn't it be "Yp= Asin(2t)+BtCos(2t)" to try and find the value of the particular solution?
  6. Jul 10, 2015 #5
    No, there are no initial conditions. The only reason I would need initial conditions any way is to solve for c1 and c2 in the initial condition. You are supposed to be able to use common coefficients in the right hand side of the equation and relate them to the unknown coefficients in the left hand side of the equation and set them equal to each other and solve. The only problem is that I need 4 equations to solve for the 4 unknowns, but I don't know how to set up the equations.

    I have shown the solution to the characteristic equation and I have solved the first half of the second solution, I split it in 2 because there are 2 functions equal to the y'' + y.

    I am only having problems finishing the second part of the second solution. Once I get the answer to Yp2 I will add it to Yp1 to make Yp. Then add Yp to my characteristic equation.

    Since the value next to the cosine is "t" that term gets a polynomial of the same degree which is (At+B) then the guess for the cosine is (Ccos(2t) + Dsin(2t)), then you multiply those together because the whole term is a t*cos(2t)
  7. Jul 10, 2015 #6
  8. Jul 10, 2015 #7


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    Instead of having two separate particular solutions, why not just have one? IOW,
    ##y_p = A\cos(2t) + B\sin(2t) + Ct\cos(2t) + Dt\sin(2t)##
  9. Jul 10, 2015 #8
    That seems like a pretty good idea, I thought of doing that but it just seemed like a hassle to go back and get rid of my work for the half of the split particular solution I already did. Also, weird considering the teacher taught us to always split up particular solutions if there are 2 terms on the right hand side.

    I will try this and get back to you
  10. Jul 10, 2015 #9

    Here is my attempt at the answer using your method...

    This is the books answer:


    I think I'm close but for some reason the coefficient pairing and solving just isn't intuitive to me...
    Is there anyone who can explain to me the process of correctly paring the coefficients and solving?
  11. Jul 10, 2015 #10


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    In your original approach, you can't write the particular solution as ##y_p = (At+B)(C\cos 2t+D\sin 2t)##. That form doesn't allow a combination of coefficients that would result in a solution with only ##t \cos 2t## and ##\sin 2t## terms, for example. To get the ##t \cos 2t## term, you'd have to have ##A \ne 0##, but then the only way to get rid of the ##t\sin 2t## term would be to set ##D=0##, which precludes the possibility of any terms having ##\sin 2t## as a factor.

    You have to use ##y_p = (At+B)\cos 2t + (Ct+D)\sin 2t##.
  12. Jul 10, 2015 #11
    Thank you, Vela, Mark explained that to me in his last post, I think I have that sorted out but now I'm having trouble solving for the A,B,C,D constants as shown in my newest uploaded work.
  13. Jul 11, 2015 #12


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    Mark's point was that using the principle of superposition to solve this problem doesn't really give you any advantage. There's no good reason to solve the problem by looking at each forcing term separately.

    Nevertheless, you can solve for a particular solution for each forcing term and combine them afterward. My point was that the form of the particular solution you tried to use for the t cos 2t forcing term was incorrect. You can't solve the problem with the trial solution you used.
  14. Jul 11, 2015 #13
    I apologize! I didn't know you were speaking in respect to the original post still, thank you, I completely understand what you are saying now, I believe.

    The only problem is still solving for those constants, once I get to that last part I'm not sure what should equal 1 and what should equal zero, and why?

    The answers I am getting with this particular arrangement is incorrect.


    P.S. This is work from the original problem where I was trying to solve t*cos(2t) when separating the two particular solutions
  15. Jul 11, 2015 #14


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    You have a sign error in your calculation of y''.
  16. Jul 12, 2015 #15


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    As a slight extra polish, you could multiply the last two terms by 3, and not change the first two since their coefficients are arbitrary.

    You ask the correct method. The correct method is the one that works and that is (for you) simplest. Ah, but if only we could always know that beforehand! Anyway, you made a guess and were told the guess is wrong, so maybe question is how to guess best.

    All I can say is that I forget the famous "methods" in between each rare time I ever solve a non-homogeneous lde. So what I do is I look at the RHS and ask myself "What is the function that (in this case) twice (in this case - in others also once or occasionally maybe three or more times) differentiated gives that?"

    Here I think I can handle the sine term so firstly the other. It's a product. Instead of cumbrously integrating it twice I think of Leibniz' Theorem for a product (uv)'' = (u''v + 2u'v' + uv'') . So say the last term of that is
    (t cos 2t) , then u is t and v is (- cos 2t)/4 . Completing the Leibniz expression I find

    [(-t cos 2t)/4]'' = sin t + t cos 2t

    On LHS I want something of form Y'' + Y. From last eq. I get:

    [(-t cos 2t)/4]'' + [(-t cos 2t)/4] = sin t + (3/4)(t cos 2t)

    but to get back the term wanted on RHS, multiply by 4/3 and get

    [(-t cos 2t)/3]'' + [(-t cos 2t)/3] = (4/3) sin t + t cos 2t. ... (1)

    This is something like the form we want, but the RHS isn't quite and calls for adding (3 - 4/3) sin 2t = (5/4) sin 2t, to get the problem's RHS. And what is (sin 2t) second derivative of?

    (- sin 2t)'' = 4 sin 2t

    of course, and hammering as before to get a form (Y'' + Y) = (5/4) sin 2t, we find

    [(-5/9) sin 2t]'' + (-5/9) sin 2t = (5/3) sin 2t. ...(2)

    Adding (1) and (2) we have the p.i. given in your text.

    Variations are possible, but doing too many steps in your head is liable to lead to mistakes that waste more time than you save.
    Last edited: Jul 13, 2015
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