What is the solution of the differential equation
`(dy)/(dx)= sec (x+y)` ?

Options

A. y=tan (x+y)=c
B. `y-c{((x+y))/2}=c`
C. `y+tan {((x+y))/2}=c`
D. `y+tan {((x-y))/2}=c`

Answer

B

Explanation

In the equation ` (dy)/(dx) = sec (x+y) `
Let x +y=v
So, ` 1+ (dy)/(dx) = (dv)/(dx) or (dy)/(dx) = (dv)/(dx) =1`
` and (dv)/(dx)-1 =secv`
` (dv)/(dx) =1+ sec v = (1 +cos v)/(cos v)`
` or,(cos v)/(1+cos v) dv = dx`
` cos v = cos^(2) v/2 -sin^(2) v/2`
` and 1 + cos v = 2 cos ^(2) v/2`
so, ` ((2 cos(2) v/2 - 2sin^(2) v/2)/(2 cos ^(2) v/2)) dv =dx`
` (1- tan^(2) v/2) dv = 2dx `
` or ,{ 1- ( sec^(2)""v/2-1))}dv = 2dx`
` or , (2 -sec^(2)""v/2)dv = 2dx`
Integrating on both the sides
` 2 int dv - int sec^(2)""v/2 dv =2 int dx +c_(1)`
where `c_(1) ` is a constant
` 2v -2tan v/2 = 2x +c_(1)`
`2(x+y) -2 tan (x+y)/2 = 2x +c_(1)`
` or, 2x +2y -2tan (x+y)/2 = 2x + c_(1)`
` y-tan ((x+y)/2) = c_(1)/2 =c`
[ c is constant]
so, ` y- tan (x+y)/2 =c`