Solve the differential equation `cos (x+y) dy=dx.` Hence find the particular solution for `x=0 and y=0.`

Explanation

The given differential equartion is
`(dy)/(dx)=(1)/(cos (x+y))" "...(1)`
Let `x+y=v`
`1+(dy)/(dx)=(dv)/(dx)`
`(dy)/(dx)=(dv)/(dx)-1`
Substituting `(dy)/(dx)` in equation (1), we get
`(dv)/(dx)-1=(1)/(cos v)`
`(dv)/(dx)=(1+cos v)/(cos v)`
`therefore` Integrating
`int(cos v dv)/(1+cos v)dv=intdx+c`
`int(1+cos v)/(1+cos v)dv-int(1)/(a+cos v)dv=intdx+c`
`int 1dv-int(1)/(2cos^(2)""(v)/(2))dv=x+c`
`v-1/2intcos^(2)""(v)/(2)dv=x+c`
`v-tan""(v)/(2)=x+c`
`(x+y)-tan((x+y)/(2))=x+c" "...(2)`
which is the required general solution of the given equation.
For `x=0 and y=0,` equation (2), becomes,
`(0+0)-tan((0)/(2))=0+c`
`impliesc=0`
Putting the value of c in equation (2), we get
`x+y-tan((x+y)/(2))=x`
`impliesy=tan ((x+y)/(2))`