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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`). |
Answer» Given differential equation is : `(x-y)(dx+dy)=dx-dy` `impliesdx+dy=(dx-dy)/(x-y)` On integration , `int(dx+dy)=int(dx-dy)/(x-y)+C` `implies x+y-log|x-y|+C`……….`(1)` Given that, when `x=0` , then `y=-1` `:. 0+(-1)=log(0+1)+CimpliesC=-1` put this value in equation `(1)`, `x+y=log|x-y|-1` `implies log|x-y|=x+y+1` which is the required particular solution. |
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