1.

Finda particular solution of the differential equation`(x" "" "y)" "(dx" "+" "dy)" "=" "dx" "" "dy`, given that `y" "=" "1`, when `x" "=" "0`. (Hint: put `x" "" "y" "=" "t`).A. ` log |x-y|= x+y +1`B. ` log |x-y|=x - y -1`C. ` log |x-y|= x+y -1`D. None of these

Answer» Correct Answer - a
Given , differential equation is
` (x-y) (dx+dy) = dx - dy`
` rArr dx + dy = (dx-dy)/(x-y)`
On integrating both sides , we get
` int (dx + dy) = int (dx-dy)/(x-y) +C`
Let `x - y = t rArr dx - dy = dt`
` :. int dx + dy -C = int (dx-dy)/(x-y) = int (dt)/t = log t = = log |x-y|`
` rArr x+ y = log + C = log | x-y | +C " "` ...(i)
It is given when x = 0 , =-1
` :. 0+(-1)= log (0+1)+C rArr C = -1 `
On substituting this value in Eq. (i) we get required prticular solution as ` x+y = log | x - y | -1 `
` rArr loh | x-y| = x+ y +1 `


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