Question

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)

Answer 1

Given: Differential equation

(x - y) (dx + dy) = dx - dy

(x - y - 1) dx + (x - y + 1) dy = 0

`therefore dy/dx = (x - y - 1)/(x - y + 1)`

Now, putting x - y = t,

`1 - dy/dx = dt/dx`

`therefore dy/dx = 1 - dt/dx`

`therefore 1 - dt/dx = (- t - 1)/(t + 1)`

or `dt/dx = 1 + (t - 1)/(t + 1)`

`= (t + 1 + t - 1)/(t  + 1)`

`=> dt/dx = (2t)/(t + 1)`

`=> dt/dx = (2t)/(t + 1)`

`=> (t + 1)/t dt = 2  dx`

On integrating,

`int (t + 1)/t dt + 2 int dx + C`

∴ `int (1 + 1/t)dt = 2x + C`

Or t + log t = 2x + C ...[putting t = x - y]

⇒ x - y + log (x - y) = 2x + C

and  log (x - y) = x + y + C

Putting x = 0, y = - 1,

0 = 0 - 1 + C

∴ C = 1

The required solution is

log(x - y) = x + y + 1