The following equation is in the form$$\frac{dy}{dx}=f(y)$$ Solve the diffrential equation$$\frac{dy}{dx}=2+3y$$
$$\frac{dy}{dx}=2+3y$$ $$dy=(2+3y)dx$$ $$\frac{dy}{2+3y}=dx$$
then I integrated bothsides.. $$\int\frac{dy}{2+3y}=\int{dx}$$
and this is what I think you should get but the book says otherwise... $$\frac{1}{3}ln(2+3y)+c = x+c$$
I know I'm wrong about this but why is there no constant when we integrate the RHS? I'm assuming there's a 1 in front of the $dx$ and when you integrate you get $x+c$
If someone could explain why I'd be grateful.