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

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