Determine whether the points (0,1) and (2,1) are on the graph of the solution of:

(t - 1) dy + 2y dt = 0, y(0) = 1.

MY WORK:

(t - 1) dy = -2y dt

-1/(2y) dy = 1/(t - 1) dt

Take integral of both sides:

-ln(|y|)/2 = ln(|t - 1|) + C

Plug in 0 for t and plug in 1 for y:

-ln(|1|)/2 = ln(|0 - 1|) + C

=> 0 = 0 + C

=> C = 0

Now I'm not sure where to go.

Thanks for the help.