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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!

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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. :)

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  • $\begingroup$ When I pondered on that answer, I came up with this crazy idea . Differential equations is a vector field, and its particular solution is the section of that certain vector field:-D $\endgroup$ – Palautot Ka May 16 '17 at 15:26
  • $\begingroup$ Geometrically one usually thinks of an ODE as a differential $1$-form, and the solutions as the induced (possibly singular) foliation by integral curves. ;) $\endgroup$ – Andrew D. Hwang May 17 '17 at 1:38
  • $\begingroup$ I will study "foliation and differential 1-form" in due timeXD.....but hey......does my conclusion "Differential equation is a vector field, and its particular solution is the section of that certain vector field" correct? $\endgroup$ – Palautot Ka May 18 '17 at 15:47
  • $\begingroup$ There's a relationship between differential equations and vector fields, but a solution is usually called an integral curve. I'm not sure what you mean by a "section of a vector field"; usually one speaks of a "section" of a surjective map$f:X \to Y$ being a map $s:Y \to X$ such that $f(s(y)) = y$ for all $y$ in $Y$. (For example, a vector field on a smooth manifold $M$ may be viewed as a section of the tangent bundle $\pi:TM \to M$.) $\endgroup$ – Andrew D. Hwang May 19 '17 at 1:30
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    $\begingroup$ I'm getting lost.....maybe I'll study differential equations in depth. The term "integral curve"? I've seen it while discussing orthogonal trajectories....ahahahahhah! Thanks man! $\endgroup$ – Palautot Ka May 19 '17 at 16:40
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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 $y$.

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  • $\begingroup$ So you mean that when we say "Get the solution of this differential equation", we really mean "Convert a differential equation to its algebraic form"? I do notice that most of the general solutions of differential equation has the form $F(x,y)=0$ or $y = f(x) + c$. Please elaborate:-) $\endgroup$ – Palautot Ka May 13 '17 at 16:43
  • $\begingroup$ @PalautotKa "Converting a differential equation to its algebraic form" I think means write down a relationship between independent and dependent variables which satisfies the relationship when you differentiate $\endgroup$ – qbert May 13 '17 at 17:03

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