Solution of the following equation
cos x dy =y(sinx-y)dx,`0ltxlt(pi)/(2)` is

Options

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

Answer

C

Explanation

We have,
`cosxdy=y(sinx-y)dx`
`rArr" "(dy)/(dx)=y tan x-y^(2)secx`
`rArr" "(1)/(y^(2))(dy)/(dx)+tanx(-(1)/(y))=-secx`
`rArr" "(dv)/(dx)+(tanx)v=secx," where v"=-(1)/(y)`
This is a linear differential equation with integrating factor
`e^(inttanxdx)=secx.` So, its solution is given by
`v sec x=-int sec^(2)xdx+C`
`rArr" "v secx=-tanx-C`
`rArr" "-(1)/(y)secx=-tanx-C`
`rArr" "secx=y(tanx+C)`
`ul("ALITER")` We have,
`cosxdy=y(sinx-y)dx`
`rArr" "cosxdy-y sinxdy=-y^(2)dx`
`rArr" "(secxdy-y tanx sec x dx)/(y^(2))=-sec^(2)xdx`
`rArr" "d((secx)/(y))=sec^(2)xdx`
`rArr" "(secx)/(y)=tanx+C`
`rArr" "secx=y(tanx+C)`