`(x-y)^(2) (dy)/(dx) =a^(2), x -y =u`

Explanation

` (x-y)^(2) (dy)/(dx) =a^(2)`
put ` x -y = u therefore x -u =y therefore 1- (du)/(dx) = (dy)/(dx) `
(1) becomes, `u^(2) (1 - (du)/(dx)) =a^(2)`
` u^(2) -u^(2) (du)/(dx) =a^(2) therefore u^(2) -a^(2) =u^(2) (du)/(dx) therefore dx= u^(2)/(u^(2) -a^(2)) du`
Integrating we get,
`int dx = int ((u^(2) -a^(2)) +a^(2))/(u^(2)-a^(2)) du+c_(1)`
`x = int 1 du + a^(2) int (du)/(u^(2)-a^(2)) +c_(1) =u +a^(2) . 1/(2a) log |(u-a)/(u+a)|+c_(1)`
` x =x -y +a/2 log |(x -y-a)/(x-y+a)|+c_(1)`
`-c_(1) + y=a/2 log |(x-y-a)/(x -y+a)|`
` -2c_(1)+2y = a log |(x -y-a)/(x-y+a)|`
` a log |(x-y-a)/(x-y+a)| =2y +c, " where " c = -2c_(1)`
This is the general solution.