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

Options

A. `tanx=(secx+c)y`
B. `secx=(tanx+c)y`
C. `ysec x=tanx+c`
D. `ytanx =secx+c`

Answer

B

Explanation

`cosxdy=y(sinx-y)dx`
`rArr (dy)/(dx)=ytanx-y^(2)secx`
`rArr 1/y^(2)(dy)/(dx)-1/ytanx=-secx`
Let `1/y=t`
`therefore -1/y^(2)(dy)/(dx)=(dt)/(dx)`
`rArr -(dt)/(dx) -t tanx=-secx`
`rArr-(dt)/(dx)+(tanx)t=secx`
I.F. `=e^(int(tanxdx))=secx`
The solution is
t(I.F.) = `int(I.F.)secxdx`
`rArr 1/ysecx=tanx+c`