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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)
Solution
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
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