Here the problem is written in lower case.
=>
Integrating we get
ln|y-2| = ln|x-1| + ln|C|
=> ln|y-2| = ln|C(x-1)|
=>| y - 2 |= |C(x-1)|
By initial condition , y = 2 when x = 2
0 = |C| i.e C = 0
Therefore the particular solution is
|y - 2| = 0
i.e. y= 2
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