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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)

Q11.     Find a particular solution of the differential equation (x - y) (dx + dy) = dx - dy, , given thaty = -1, when x = 0. (Hint: put x - y = t)

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Given equation is
(x - y) (dx + dy) = dx - dy,
Now, integrate both the sides
Put
(x-y ) = t\\ dx - dy = dt
Now, given equation become
dx+dy= \frac{dt}{t}
Now, integrate both the sides
x+ y + C '= \log t + C''
Put t = x- y  again
x+y = \log (x-y)+ C
Now, by using boundary conditions we will find the value of C
It is given that
y = -1 when x = 0
0+(-1) = \log (0-(-1))+ C\\ C = -1
Now, put the value of C

x+y = \log |x-y|-1\\ \log|x-y|= x+y+1
Therefore,  the particular solution of the differential equation (x - y) (dx + dy) = dx - dy,   is   \log|x-y|= x+y+1

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