# How can I understand the solution of an ordinary differential equation?

I'm reading a textbook and it is talking about the solution to a differential equation $$P(x,y) \text{d}x +Q(x,y)\text{d}y = 0$$. But I find it hard to understand this equation. What does it mean? And what is defined to be a solution of the equation?

For example, my textbook claims that for an equation of the form $$X(x)F(y)\text{d}x + Y(y)G(x) \text{d}y = 0$$, if $$G(a)=0$$ then " the function $$x = a$$ is a solution to the equation." I really don't understand. Can anyone help ?

$$dx$$ and $$dy$$ are differentials, which can be understood as relative displacements along the tangent line to a curve. So if you start out at a point $$(x,y)$$ on the curve and move an arbitrary distance along the tangent line, if the $$x$$ coordinate changes by $$dx$$ and the $$y$$ coordinate changes by $$dy$$, then the differential equation says $$P(x,y) dx + Q(x,y) dy$$ should be $$0$$. If that is true for all points on the curve, the curve is a solution of the differential equation. If you consider $$x$$ to be the independent variable and $$y$$ the dependent variable, so that on the curve $$y$$ is a certain function of $$x$$, then $$\dfrac{dy}{dx}$$ is the usual derivative of $$y$$ with respect to $$x$$, and the differential equation says $$P(x,y) + Q(x,y) \dfrac{dy}{dx} = 0$$. But might just as well think of $$x$$ as the dependent variable and $$y$$ the independent variable, and then the differential equation says $$P(x,y) \dfrac{dx}{dy} + Q(x,y) = 0$$. It's the same curve, but looked at from different directions ($$y$$ as a function of $$x$$, or $$x$$ as a function of $$y$$).
In your example, $$dx =0$$, $$x=a$$ is constant and the tangent line is vertical, so this one doesn't work with $$x$$ as independent variable, only with $$y$$ as independent variable. In these terms the differential equation is $$X(x) F(y) \dfrac{dx}{dy} + Y(y) G(x) = 0$$ which is satisfied when $$x$$ is the constant function $$a$$: the first term is $$0$$ because $$dx/dy = 0$$ and the second term is $$0$$ because $$G(a)=0$$.