The solution of differential equation
` cos x dy = y (sin x - y ) dx, 0 lt x lt pi //2 ` is

Options

A. `sec x = (tan x + C)y`
B. `y sec x = tan x +C`
C. ` y tan x = sec x +C`
D. ` tan x = (sec x + C)y`

Answer

a

Explanation

Since , ` cos x dy = y sinx dx - y^(2)dx`
` rArr 1/(y^(2))(dy)/(dx) -1/y = (dz)/(dx)`
` rArr (dz)/(dx) +(tanx) z = - sec x`
This is a linear differential equuation . Therefore
`IF = e^(int tanx dx) = e^(log(secx)) = sec x`
Hence, the solution is
` z* (sec x) = int - sec x * sec x dx +C_(1)`
` rArr-1/y secx = - tan x + C_(1)`
` rArr sec x = y(tan x +C) " " [ C = -C_(1)]`