MAT244-2013S > MidTerm
MT Problem 2b
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
Navigation
[0] Message Index
[#] Next page
Go to full version