cos (x+y) dy =dx , when x =0 and y =0

Explanation

cos (x+y) dy =dx
` therefore (dy)/(dx) = 1/(cos (x+y))`
Put x +y =v, then ` 1 + (dy)/(dx) =(dv)/(dx)`
`(dy)/(dx) = (dv)/(dx) -1`
(1) becomes, ` (dv)/(dx) -1 = 1/(cos v) " " therefore (dv)/(dx) = 1/(cos v) + (1 +cos v)/(cos v)`
` (cos v)/(1+ cos v) dv =dx`
Integrating we get,
` int (cos v)/( 1 + cos v) dv = int dx +c`
`int (1+ cos v -1)/(1 + cos v) dv = int dx +c`
`int (1- 1/(2 cos^(2) (v/2))) dv= int dx +c `
` int 1 dv -1/2 int sec^(2) (v/2)dv = int dx +c`
` v -1/2 (tan(v/2))/((1/2)) = x+c`
` x+y -tan ((x+y)/2) = x +c`
` y = tan ((x+y)/2)+c`
This is the general solution.
When x =0 , y=0 , we get,
0 = tan +c ` therefore ` c=0
The particular solution is y= ` tan ((x+y)/2)`