Solve the following differential equations with the help of the subsitutions shown against then :
` (dy)/(dx) =cos (x+y), x + y = v `

Explanation

` (dy)/(dx) = cos (x+y) `
Put x + y = v , ` " " therefore 1 + (dy)/(dx) = (dv)/(dx) `
` (dy)/(dx) = (dv)/(dx) -1`
(1) becomes, ` (dv)/(dx) -1 = cos v`
`(dv)/(dx) = 1 + cos v `
`(dv)/(dx) =1 + cos v " " therefore 1/(1 + cos v) dv = dx`
INntegrating , we get
` int 1/(1 + cos v) dv = int dx + c`
` int 1/(2 cos^(2) (v/2)) dv = int dx + c`
` 1/2 int sec^(2) (v/2) dv =int dx +c`
`1/2 . (tan (v/2))/((1/2)) = x+c " " therefore ((x+y)/2) = x + c`
THis is the general solution .