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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 that$y = -1$, 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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