Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0. (Hint: put x – y = t)


It is given that (x – y)(dx +dy) = dx – dy

(x –y + 1)dy = (1- x + y)dx



----------------(1)


Let x – y = t





Now, let us substitute the value of x-y and in equation (1), we get,







-------(2)


On integrating both side, we get,


t + log|t| = 2x + C


( x – y ) + log |x – y| = 2x + C


log|x – y| = x + y + C --------(3)


Now, y = -1 at x = 0


Then, equation (3), we get,


log 1 = 0 -1 + C


C = 1


Substituting C = 1in equation (3), we get,


log|x – y| = x + y + 1


Therefore, a particular solution of the given differential equation is log|x – y| = x + y + 1.


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