Solve `2(y+3)-x y(dy)/(dx)=0`, given that `y(1)=-2`

Explanation

Given that, `" "2(y-3)-xy(dy)/(dx)=0`
`rArr" "2(y+3)=xy(dy)/(dx)`
`rArr" "2(dx)/(x)=((y)/(y+3))dy`
`rArr" "2*(dx)/(x)=((y+3-3)/(y+3))dy`
`rArr" "2*(dx)/(x)=(1-(3)/(y+3))dy`
On integrating both sides, we get
`" "2logx=y-3log(y+3)+C" "...(i)`
When x= and y=-2, then
`" "2log1=-2-3log (-2+3)+C`
`rArr" "2*0=-2-3*0+C`
`rArr" "C=2`
On subsituting the value of C in Eq. (i), we get
`" "2logx=y-3log(y-3)+2`
`rArr" "2logx+3log(y+3)=y+2`
`rArr" "logx^(2)+log(y+3)^(3)=(y+2)`
`rArr" "logx^(2)(y+3)^(3)=(y+2)`
`rArr" "x^(2)(y+3)^(3)=e^(y+2)`