What is the solution of the difrerential equation (x+y) (dx -dy) =dx+dy ?

Options

A. 2log (x+y)=c (y-x)
B. (y-x)+log (x+y) =c
C. `(y/x)+[log(y/x)]=c`
D. None of these

Answer

B

Explanation

Given differenital equation is :
` (x+y) (dx-dy) = dx +dy `
` Rightarrow (x+y) dx -(x+y) dy = dx +dy `
` Rightarrow (x+y -1) dx = (x +y +1) dy`
` Rightarrow (dy)/(dx) = (x +y -1)/(x+y+1)`
Let x +y and ` (dy)/(dx) = (dv)/(dx)-1`
` (dv)/(dx) -1 = (v -1)/( v+1)`
` Rightarrow (dv)/(dx) = (v-1)/(v+1)+1`
` Rightarrow (dv)/(dx) = ( v-1 +v +1)/(v+1)`
` Rightarrow (v+1)/(2v) dv = dx`
` Rightarrow 1/2int 1dv + 1/2 int 1/v dv = int 1dx`
` Rightarrow 1/2 v + 1/2 log v =x +c_(1)`
` Rightarrow +y+log (x+y) = 2x +c " " (therefore 2c_(1) =c= " costant ")`
` Rightarrow (y-x) + log (x +y) =c`