Question

What is a solution to the differential equation `dy/dx=x/(y+3)`?

Answer 1

`y =pm sqrt( x^2 + C) - 3`

Explanation:

this is separable

`y' = x/(y+3)`

`(y+3) y'= x`

`int \ (y+3) y' \ dx =int \ x \ dx`

See ASIDE below
`implies int \ y+3 \ dy =int \ x \ dx`

`y^2/2+3y = x^2/2 + C`

`y^2+6y = x^2 + C`

completing the square
`(y+3)^2 - 9 = x^2 + C`

`(y+3)^2 = x^2 + C`

`y =pm sqrt( x^2 + C) - 3`

ASIDE
Avoiding doing algebra with the expression `dy/dx`
Because `I = int \ (y+3) y' \ dx`
then `d/dx (I) = (y+3) y'`
And from the chain rule and the fact that `y = y(x)` we can say that
`d/dx( I ) = I' y'`
so `I' = y+3`
`implies I = int \ y+3 \ dy`