### Author Topic: MT Problem 2a  (Read 3621 times)

#### Victor Ivrii

• Elder Member
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##### 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
##### Re: MT Problem 2a
« Reply #1 on: March 06, 2013, 10:15:48 PM »
part 1

#### Victor Ivrii

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.