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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