C=0, so,

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

ln|y^(-1/2)| = ln|t-1|

So,

|y^(-1/2)| = |t -1|

|1/(y^(1/2))| = |t-1|

Square both sides,

1/y = (t-1)^2 ----------(i)

Test (0,1) on (i),

1/1 =? (0-1)^2

1 =? 1

Yes, so, (0,1) is on the graph of (i).

Test (2,1) on (i),

1/1 =? (2-1)^2

1 =? 1

Yes, so, (2,1) is also on the graph of (i).