Convert a differential equation into an algebraic equation? The book I used (Calculus with Analytic Geometry by Thurman S. Peterson (printed in 1960)) says that:
"A relation among the variables which reduces a differential equation to an algebraic identity is called a solution of the equation."
Does it mean that we just convert a differential equation into its algebraic equation form and to verify that this algebraic equation comes from a certain differential equation, we called this algebraic equation the solution to a certain differential equation?
Seems that the word "solution" in differential equations is not a traditional one. I wanna be enlightened. Thanks!
 A: To elaborate on qbert's answer, a (first-order) "differential equation" is often defined to be an algebraic equation in $x$, $y$, $dx$, and $dy$ formally equivalent to $P(x, y)\, dx + Q(x, y)\, dy = 0$ for some continuous functions $P$ and $Q$ defined throughout some plane region $R$. Examples include
$$
\frac{dy}{dx} = y;\quad
x\, dx + y\, dy = 0;\qquad
\frac{y\, dx - x\, dy}{x^{2} + y^{2}} = 0.
$$
The (general) "solution" is a continuously-differentiable function $F$ defined in $R$ for which the level curves $F(x, y) = C$ "satisfy the differential equation" in the sense that implicit differentiation of $F(x, y) = C$ yields a condition formally equivalent to $P(x, y)\, dx + Q(x, y)\, dy = 0$. Solutions to the three preceding examples might look like
$$
ye^{-x} = C;\qquad
x^{2} + y^{2} = C;\qquad
\frac{y}{x} = C.
$$
Notes:


*

*The term "algebraic identity" refers to the absence of $dx$ and $dy$; that is, "algebraic" contrasts with "differential", not (say) with "transcendental".

*The function $F$ in a solution is far from unique: If $g$ is any strictly monotone function of one variable, for example, then $(g \circ F)(x, y) = C$ is another way of expressing the solutions of a first-order differential equation. Thus
$$
\log(1 + x^{2} + y^{2}) = C;\qquad
e^{-(x^{2} + y^{2})} = C;
$$
are two other ways of expressing $x^{2} + y^{2} = C$. (Of course, the meaning of $C$ is not the same in these three representations.)

*"Just convert a differential equation into its algebraic equation form" makes the process sound algorithmic, even trivial, which generally it is not. :)
A: I think what is going on here is this definition allows for implicit solutions; i.e. some curve 
$$
F(x,y)=0
$$
where $y$ is dependent and $x$ is independent. Such a curve may not be solvable for $y$ but it is an algebraic identity in $x$ and $y$.
