Author Topic: Particular Solution Question  (Read 667 times)

Chang Peng (Eddie) Liu

  • Full Member
  • ***
  • Posts: 19
  • Karma: 7
    • View Profile
Particular Solution Question
« on: December 06, 2014, 01:34:45 PM »
$$y^{(4)}−3y″−4y=\sin(t)+8t.$$

apparently the particular solution is $Y = At\cos(t) + Btsin(t) + Ct + D$

I know why it's $At\cos(t) + Bt\sin(t)$ , but I have no idea where the $Ct + D$ is coming from. Can someone please explain this?
« Last Edit: December 06, 2014, 05:47:07 PM by Victor Ivrii »

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Particular Solution Question
« Reply #1 on: December 06, 2014, 05:46:51 PM »
Observe that $r_{1,2}=\pm 2$, $r_{3,4}=\pm i$.

Because right-hand expression is $f_1+f_2$ with $f_1=\sin(t)$ and $f_2=8t$. Then we need $Y=Y_1+Y_2$ with  $Y_1= (A\cos(t)+B\sin (t))t $ and $Y_2= Ct+D$.
« Last Edit: December 06, 2014, 11:51:36 PM by Victor Ivrii »

Chang Peng (Eddie) Liu

  • Full Member
  • ***
  • Posts: 19
  • Karma: 7
    • View Profile
Re: Particular Solution Question
« Reply #2 on: December 06, 2014, 08:08:35 PM »
That makes sense. Thank you professor.

Sheng Zang

  • Jr. Member
  • **
  • Posts: 5
  • Karma: 0
    • View Profile
Re: Particular Solution Question
« Reply #3 on: December 06, 2014, 09:55:50 PM »
Prof, why you make the particular solution Y1=(Acos(t)+Bsin(t))t^2?
I think for this question, the particular solution multiplies by t is enough, i.e Y1= Atcos(t)+Btsin(t).
Thank you.

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Particular Solution Question
« Reply #4 on: December 06, 2014, 11:52:00 PM »
Yes, it was a misprint. Corrected