MAT244-2013S > MidTerm

MT Problem 2b

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Victor Ivrii:
Find a particular solution of equation
\begin{equation*}
(t^2-1) y''-2ty'+2 y=1.
\end{equation*}
Hint: use variation of parameters.

Brian Bi:
By inspection, $y = 1/2$ is a solution.

Patrick Guo:
The homogeneous sol'ns were y1=t, y2=t2 right?

But why Wolfram alpha gives y(t) = (c_1 sqrt(t^2-1) (1-t)^(3/2))/sqrt(t+1)+(c_2 t sqrt(t^2-1) sqrt(1-t))/((t-1) sqrt(t+1))
...

the 2nd term of Wolfram is equivalent {if regardless of the signs (consider all things in roots being positive)} ; but I have no idea what the first term is ... hold on, it can be written as    constant *(t-1)^2    , so it's still correct   :P

Patrick Guo:
$$y(t) =(c_1 \sqrt{t^2-1} (1-t)^{3/2})/\sqrt{t+1}+(c_2 t \sqrt{t^2-1} \sqrt{1-t})/((t-1) \sqrt{t+1})$$

[Don't know how to get the codes working..]

Jeong Yeon Yook:
solution