The solution of equation `(2y-1)dx-(2x+3)dy=0` is

Options

A. `(2x-1)/(2y+3)=k`
B. `(2y+1)/(2x-3)=k`
C. `(2x+3)/(2y-1)=k`
D. `(2x-1)/(2y-1)=k`

Answer

C

Explanation

Given that, `" "(2y-1)dx-(2x+3)dy=0`
`rArr" "(2y-1)dx=(2x+3)dy`
`rArr" "(dx)/(2x+3)=(dy)/(2y-1)`
On integrating both sides, we get
`" "(1)/(2) log(2x+3)=(1)/(2)log(2y-1)+logC`
`rArr" "(1)/(2)[log*(2x+3)-log(2y-1)]=logC`
`rArr" "(1)/(2)log((2x+3)/(2y-1))=logC`
`" "((2x+3)/(2y-1))^(1//2)=C`
`rArr" "(2x+3)/(2y-1)=C^(2)`
`rArr" "(2x+3)/(2y-1)=k`, where `K=C^(2)`