Processing math: 0%

Author Topic: Problem 1 (noon)  (Read 16995 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2607
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Problem 1 (noon)
« on: November 19, 2019, 04:15:32 AM »
(a) Find the general solution of
y

(b) Find solution satisfying
y(0)=y'(0)=0.

Yiheng Bian

  • Full Member
  • ***
  • Posts: 29
  • Karma: 12
    • View Profile
Re: Problem 1 (noon)
« Reply #1 on: November 19, 2019, 04:33:17 AM »
No double-dipping

(a):
We can solve homo firstly:
r^2-3r+2=0\\ (r-2)(r-1)=0\\ r_1=2,r_2=1
Therefore:
y=c_1e^{2t}+c_2e^t
So we can get:
W=\begin{vmatrix} e^{2t} & e^t \\ 2e^{2t} & e^t \end{vmatrix}=-e^{3t}\\ W_1=\begin{vmatrix} 0 & e^{t} \\ 1 & e^{t} \end{vmatrix}=-e^{t}\\ W_2=\begin{vmatrix} e^{2t} & 0 \\ 2e^{2t} & 1 \end{vmatrix}=e^{2t}
So we can get:
Y(t)=e^{2t}\int{\frac{-e^{s}*\frac{e^{3s}}{e^{s2}+1}}{-e^{3s}}}ds + e^{t}\int{\frac{e^{2s}*\frac{e^{3s}}{e^{s2}+1}}{-e^{3s}}}ds\\ Y(t)=e^{2t}\int{\frac{e^{s}}{e^{2s}+1}}ds - e^{t}\int{\frac{e^{2s}}{e^{2s}+1}}ds\\ Y(t)=e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)
Finally:
y(t)=c_1e^{2t}+c_2e^t+e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)




(b):
So we can get y'(t):
y'=2c_1e^{2t}+c_2e^t+2e^{2t}arctan(e^t)+e^{2t}*\frac{e^t}{e^{2t}+1}-0.5e^t*ln(e^{2t}+1)-0.5e^t*\frac{e^{2t}}{e^{2t}+1}
We take y(0)=y'(0)=0,so we can get:
2c_1+2c_2+0.5\pi-ln2=0\\ 2c_1+c_2+0.5\pi-0.5ln2=0
So
c_1=-0.25\pi,c_2=0.5ln2
Therefore:
y=-0.25\pi*e^{2t}+0.5ln2e^t+e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)
« Last Edit: November 24, 2019, 08:32:06 AM by Victor Ivrii »

NANAC

  • Jr. Member
  • **
  • Posts: 10
  • Karma: 2
    • View Profile
Re: Problem 1 (noon)
« Reply #2 on: November 19, 2019, 09:04:42 AM »
Please see the attachment for the answer

OK.

But No snapshots!
« Last Edit: November 24, 2019, 08:39:48 AM by Victor Ivrii »

xilin zhang

  • Jr. Member
  • **
  • Posts: 5
  • Karma: 2
    • View Profile
Re: Problem 1 (noon)
« Reply #3 on: November 19, 2019, 09:11:19 AM »
I got a different y' in part b.

baixiaox

  • Jr. Member
  • **
  • Posts: 10
  • Karma: 0
    • View Profile
Re: Problem 1 (noon)
« Reply #4 on: November 19, 2019, 05:34:40 PM »
Answer for question1

Mingdi Xie

  • Jr. Member
  • **
  • Posts: 14
  • Karma: 0
    • View Profile
Re: Problem 1 (noon)
« Reply #5 on: November 20, 2019, 02:29:07 PM »
This is my solution

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2607
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Problem 1 (noon)
« Reply #6 on: November 24, 2019, 08:41:08 AM »
\boxed{  y= \Bigl(-\frac{1}{2}\ln (e^{2t}+1)+c_1 \Bigr)e^{t} + \Bigl( \arctan (e^t)+c_2\Bigr)e^{2t}. }
 and
\boxed{  y= \Bigl(-\frac{1}{2}\ln (e^{2t}+1)+\frac{1}{2}\ln (2)  \Bigr)e^{t} + \Bigl( \arctan (e^t)-\frac{\pi}{4}\Bigr)e^{2t}. }