Solution of the differential equation
`cosxdy=y(sinx-y)dx, 0ltxlt(pi)/(2)`is

Options

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

Answer

A

Explanation

since `cosxdy=ysinx`
`dx-y^(2)dx`
`implies(1)/(y^(2))(dy)/(dx)-(1)/(y)tanx=-secx`
Put, `-(1)/(y)=zimplies(1)/(y^(2))(dy)/(dx)=(dz)/(dx)`
`implies(dz)/(dx)+(tanx)z=-secx`
This is a linear differential equation. Therefore
`lf=e^(int tanxdx)=e^(logsecx)=sex`
Hence, the solution is
`z(secx)=int-secxsecxdx+c_(1)`
`implies-(1)/(y)secx=-tanx+c_(1)`
`implies secx=y(tanx+c)`