The solution of the differential equation `dy/dx=(x-y)^2," when " y(1)=1`, is

Options

A. `-log _(e) |(1+x-y)/(1-x+y)|=x+y-2`
B. `log _(e) |(2-y)/(2-x)|=2 (y-1)`
C. `-log _(e) |( 1-x +y)/(1+ x-y)|=2 (x-1)`
D. `log _(e) |(2-x)/(2-y) |=x-y`

Answer

C

Explanation

`because (dy)/(dx) = (x-y) ^(2) " "…(i)`
Put `x -y =1implies 1-(dy)/(dx) =(dt)/(dx )implies (dy)/(dx ) =1 - (dt)/(dx) therefore` From (i) `1- (dt)/(dx) =t ^(2) implies 1-t ^(2) = (dt)/(dx) implies (dt)/(1-t ^(2)) =dx`
Integrating, `int (dt)/(1-t ^(2)) = int dx +C`
`1/2 ln |(1+t)/(1-t)| =x +C implies 1/2 ln (|1 + x -y)/(1-x + y|) =x +C " "...(ii)`
Given that `y (1) =1 implies C =-1`
`therefore` From (ii)
`1/2 ln |(1+x-y)/(1-x+y)|=x-1implies-ln |(1-x+y)/(1+x-y)|=2 (x -1)`