The general solution of differential equation `(dx)/(dy)=cos(x+y)` is

Options

A. `tan((x+y)/(2))=y+C`
B. `tan((x+y)/(2))=x+C`
C. `cot((x+y)/(2))=y+C`
D. `cot((x+y)/(2))=x+C`

Answer

A

Explanation

We have,
`(dx)/(dy)=cos(x+y)`
Put `x+y=v`
`implies(dx)/(dy)+1=(dv)/(dy)`
`implies(dx)/(dy)=(dv)/(dy)-1`
`therefore` Equation becomes ,brgt `(dv)/(dy)-1=cosv`
`implies(dv)/(dy)=1+cosv`
`implies(dv)/(1+cosv)+dy`
`implies(dv)/(2"cos"^(2)(v)/(2))=dy`
On integrating, we get
`(1)/(2)int"sec"^(2)(v)/(2)dv=intdy`
`implies"tan"(v)/(2)=y+C`
`impliestan((x+y)/(2))=y+C" "[becausev=x+y]`