Author Topic: MT Problem 2a  (Read 1190 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
MT Problem 2a
« on: March 06, 2013, 09:06:36 PM »
Find solution $y_2(t)$ of
\begin{equation*}
(t^2-1) y''-2ty'+2 y=0
\end{equation*}
where one of the solutions is $y_1(t)=t$ and solution $y_2$ is such that  $W(y_1,y_2)=-1$ at $t=0$ and $y_2(0)=1$.

Jeong Yeon Yook

  • Full Member
  • ***
  • Posts: 20
  • Karma: 8
    • View Profile
Re: MT Problem 2a
« Reply #1 on: March 06, 2013, 10:15:48 PM »
part 1

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: MT Problem 2a
« Reply #2 on: March 07, 2013, 03:45:20 AM »
J. Yook' solution is correct. Note $\int \frac{2t}{t^2-1} \,dt = \ln (1-t^2)+ \ln C$ as numerator is a derivative of denominator (no need to invoke partial fractions). It should be also noted that we are talking about interval $(-1,1)$ as $t=\pm 1$ are breaking points (senior coefficient vanishes there) and $t=0$ selects this interval.