The solution of the differential equation `dy/dx + y/2 secx = tanx/(2y),` where `0 <= x < pi/2,` and `y(0)=1,` is given by

Options

A. `y^(2)=1-(x)/(sec x +tanx)`
B. `y^(2)=1+(x)/(sec x +tan x)`
C. `y=1+(x)/(sec x +tan x)`
D. `y=1-(x)/(sec x+tan x)`

Answer

A

Explanation

We have,
`(dy)/(dx)+(1)/(x)y sec x =(tanx)/(2y)`
`rArr" "2y(dy)/(dx)+y^(2)sec x=tanx`
`rArr" "(du)/(dx)+(sec x)u=tan x, " where u"=y^(2)`
This is a linear differential equation with
`"I.F."=e^(int secxdx)=e^(log(secx+tanx))=secx+tanx.`
Multiplying both sides by I.F. `=sec x+tanx` and integrating we get
`u(secx+tanx)=inttanx(secx+tanx)dx+C`
or, `y^(2)(secx+tanx)=secx+tanx-x+C" ...(i)"`
It is given y = 1 when x = 0. Putting x = 0, y = 1 in (i) we get
C = 0. Putting C = 0 in (i), we obtain
`y^(2)(secx+tanx)=sec x+tanx-x`
`rArr" "y^(2)=1-(x)/(sec x+tanx)`