1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differential equation: undetermined coeff.

  1. Mar 22, 2007 #1
    1. The problem statement, all variables and given/known data
    i'm stuck. could someone please help me with this? thanks!
    find the general solution:

    y''+2y'+y=2e^-t

    2. Relevant equations

    CE: r^2+2r+1=0
    (r+1)^2 which means r1=r2=-1
    c1e^-t+c2te^-t

    3. The attempt at a solution

    Y(t)=Ae^-t
    Y'(t)=-Ae^-t
    Y''(t)=Ae^-t
    back into the equation:
    Ae^-t-2Ae^-t+Ae^-t=2Ae^-t
    0=Ae^-t
    now im stuck! what am i doing wrong?! this is the "easiest" problem! isn't the A left there and that's how i find the GS?

    the answer is supposed to be c1e^-t+c2tAe^-t+t^2Ae^-t
     
  2. jcsd
  3. Mar 22, 2007 #2
    Have you figured out the homogeneous solution yet? Do that and see if your problem pops out. (Hint: What do you think will happen if you guess the same form as your homogeneous solution?)
     
  4. Mar 22, 2007 #3
    but isn't your guess supposed to be Ae^-t because that's what fits. sorry. but not sure if i'm getting your question. thanks for the input.
     
  5. Mar 22, 2007 #4
    Yes, typically that would be a good guess at the particular solution, but essentially what happens is that the solution space (general solution) is the superposition of the homogeneous solution and the particular solution. If you add a constant multiple of the homogeneous solution then that will get you nowhere. In other words, if your guess at the particular solution has the same form as the homogeneous solution then you will not find the other linear function in the superposition of the general solution because L1[x] + cL1[x] doesn't really mean anything.
     
  6. Mar 22, 2007 #5
    ok i tried Ae^-t+Bte^-t as a guess. but it all canceled out still.
     
  7. Mar 23, 2007 #6
    Of course it did. The homogeneous solution is c1e^-t + c2te^-t. Your guess is exactly the same.

    You need to try another.
     
  8. Mar 23, 2007 #7
    can i have a hint? =) i just tried Ate^-t from another book that says if the assumed solution is a multiplicative function of the homogenous solution and that it should be multiplied by t^m where m is the smallest integer possible. (did i do that right?)
     
  9. Mar 23, 2007 #8
    Alright, you were on the right track by continually adding coefficients and higher orders of t.

    Did you know that you could solve an equation that only needed Ae^-t with as many higher order terms as you wanted?
    Zt^n e^-t + Yt^n-1 e^-t + ... + Bt e^-t + Ae^-t would give the same thing for an equation of say y'' + y = 10e^-t as guessing the lowest order term Ae^-t.
     
  10. Mar 23, 2007 #9
    ok, so i tried this:

    Ate^-t+Bt^2 e^-t+C (not sure if i need the C)

    but after all the algebra, B=1, and i'm not sure how to find A. am i on the right track? based on your explanation....i might be...

    ahhhh i tried this like 6 times...
     
    Last edited: Mar 23, 2007
  11. Mar 23, 2007 #10
    wait. A=C=0? so the solution is homogenous solution+this so it is:

    c1e^-t+c2te^-t+t^2e^-t since B=1 -> 1te^-t?
     
  12. Mar 23, 2007 #11
    Yep, you got it. It seems like you are not really sure why A and C are zero though.

    So if you guess the particular solution
    [tex]y_p(t) = Ae^-t + Bte^-t + Ct^2e^-t[/tex]

    You will find a huge mess in your particular solutions. After a lot of algebra, which you have to be careful with and really just take it slow, you will find that

    [tex] 2e^-t = 2Ce^-t[/tex]

    multiply the equation by e^t and C = 1.

    The other coefficients are completely gone, and so they must be zero.

    Thus, the particular solution will be [tex]y_p(t) = 0e^-t + 0te^-t + 1t^2 e^-t [/tex].

    The general solution is the superposition of the homogeneous and particular.
     
  13. Mar 23, 2007 #12
    Ah crap, I forgot brackets. Just note that every time you see a e^-*t I really meant to do e^(-t). Don't really want to fix it. :)
     
  14. Mar 23, 2007 #13
    ok great, thanks! you've been a lot of help. but i guess i still made a mistake because my guess was Ate^-t+Bt^2 e^-t+C instead of Ae^-t+Bte^-t+Ct^2e^t. so just to confirm, the guess should be At^-t + Bt^-t + Ct^2e^-t? i thought it was just to multiply the homogeneous solution by a factor of t...sorry...taking a while to digest, this stuff is still really new.
     
  15. Mar 23, 2007 #14
    Yeah, you should have had my guess, which is really (A + Bt + Ct^2)e^-t if that helps you see it a little clearer. You just have to make sure your guess at the particular solution doesn't copy the homogeneous solution because it will not have the range (solution space) as the homogeneous.

    Say you had a homogeneous solution of c1e^t + c2t^2e^t + c3t^3e^t + c4t^4e^t then your particular would have to have one higher order than the homogeneous y_p = (A + Bt +Ct^2 + Dt^3 + Et^4 + Ft^5)e^-t.
     
  16. Mar 23, 2007 #15

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You don't need all of (A+ Bt+ Ct2)e-t. Since both e-t and te-t are solutions to your homogeneous equation, try y= At2e-t.
     
  17. Mar 23, 2007 #16
    great, thanks everyone. ok i'm just practicing problems now since i have a midterm coming up. i'm using schaum's outlines: differential equations. here's the problem:

    y''-2y+y=t^2-1

    CE: (r-1)^2, r1=r2=1
    yh=c1e^t+c2te^t

    GS=yh+yp

    my guess: yp=At^2+Bt+C
    yp'=2At+B
    yp''=2A

    (question: since it's a repeated root, should i multiply my guess by a factor of t?)

    so after all the algebra, i got yp=t^2-4t-11

    the book solution is yp=t^2+4t+5; GS=c1*e^t+c2*te^t+t^2+4t+5

    did i make an algebra mistake? i went through and looked, it doesn't look like it. did i choose my guess wrong? why?
     
  18. Mar 23, 2007 #17

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I assume you mean y"- 2y'+ y= t^2- 1

    Since your solution to the entire equation does not involve et, no.

    yp"= 2A
    yp'= 2At+ B so -2yp'= -4At- 2B
    yp= At2+ Bt+ C
    yp"- 2yp'+ yp= 2A- 4At- 2B+ At2+ Bt+ C
    = At2+ (B-4A)t+ (2A- 2B+ C)= t2- 1
    so we must have A= 1, B- 4A= 0, 2A- 2B+ C= -1.
    B- 4(1)= 0 so B= 4, 2(1)-2(4)+ C= -1 so C= -1+ 8- 2= 5

    I suspect that you lost track of signs!

     
    Last edited: Mar 26, 2007
  19. Mar 24, 2007 #18
    i just realized. for the first problem, my professor said resonance analysis must be performed. how do you know when you need to do this?
     
  20. Mar 24, 2007 #19
    What do you, and your professor, mean by resonance analysis?
     
  21. Mar 24, 2007 #20
    umm well he said resonance analysis is make r = alpha + beta*i s=# of times r appears as a root of CE. the equation he keeps using is:

    y=t^s * e^(alpha*t) * (An * cos(beta*t) + Bn * sin(beta*t))

    i'm not sure when to apply this, but one of my classmates said i am supposed to perform this analysis to get Ate^-t as a guess
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Differential equation: undetermined coeff.
  1. Undetermined Coeff. (Replies: 1)

Loading...