Solve the differential equation
`"cosec"x logy(dy)/(dx) + x^(2)y^(2)=0.`

Explanation

The given differential equation may be written as
`"cosec"x logy(dy)/(dx) =- x^(2)y^(2)`
` rArr (logy)/(y^(2))dy = -x^(2) sinx dx " " `[separating the variables]
`rArr int underset(I)((log y))((1)/(underset(II)(y^(2))))dy = -int underset (I)(x^(2)) underset(II)(sinx) dx " " `[integrating both sides].
Integrating by parts on each side, we get:
` (logy) ((-1)/(y)) -int (1)/(y) *((-1)/(y)) dy = -[x^(2) (-cos x) - int 2x(-cos x)dx ] +C_(1)," where " C_(1)` is an arbitrary constant
`rArr (-logy)/(y) +int(1)/(y^(2))dy =x^(2) cosx -2int underset (I)(x)underset(II)(cos x) dx +C_(1)`
`rArr (-log y)/(y) -(1)/(y) =x^(2)cosx -2x sinx +2 int sinxdx +C_(1)` [integrating by parts]
`rArr (-log y)/(y) -(1)/(y) =x^(2) cosx -2x sinx -2 cosx +C_(1)`
`rArr (logy)/(y)+(1)/(y) = -x^(2) cosx +2x sinx +2 cos x +C, " where " -C_(1) =C.`
This is the required solution of the given differential equation.