# When we write $y'' + y' + y = x$, for a function $y$, do we really mean $y''(x) + y'(x) + y(x) = x$?

I have recently started exploring differential equations at a graduate level (partial and ordinary) and really started thinking about why we write them the way we do. Let (since this seems to be the common symbol of choice) $$y: \mathbb{R} \to \mathbb{R}$$ such that $$x \mapsto y(x)$$. Just to be fussy, we will consider $$y \in C^{\infty}(\mathbb{R})$$. Now, for example, consider the second order differential equation $$y'' + y' + y = x \tag{i}$$ My question is in regards to this notation: Why do we write the function and it's derivatives this way in the equation rather than writing: $$y''(x) + y'(x) + y(x) = x \tag{ii}$$ using the function values? When we write (i), do we really mean (ii)? This may be a bit pedantic, but I just want to be thorough.

• Yep, that is what we mean and we do it because it is shorter. It is annoying in the beginning, but when you got used to it, it is slicker and easier to grasp. Dec 29, 2020 at 15:15
• I mean, it is much nicer to look at the equation $$u_{xxx} - 6u u_x = u_t$$ than $$u_{xxx}(x,t) - 6 u(x,t) u_{x}(x,t) = u_t (x,t).$$ Writing the variables does not add any additional information. Dec 29, 2020 at 15:18
• We should actually write (i) as $y''+y'+y=\text{id}$, but somehow we never do. ;-) Dec 29, 2020 at 15:31
• @SeverinSchraven - Thats very helpful, thank you! Dec 29, 2020 at 19:31

## 1 Answer

The context of the equation $$$$y^{\prime\prime} + y^{\prime} + y = x$$$$ means that given a function described by the following proposition: $$$$\forall x \in \mathbb{R}, y^{\prime\prime}\left(x\right) + y^{\prime}\left(x\right) + y\left(x\right) = x,$$$$ use deduction rules to deduce a new proposition in the following form: $$$$\forall x \in \mathbb{R}, y\left(x\right) = ...$$$$ Of course, rigorously speaking, you should first prove the existence and uniqueness of such a function $$y$$: $$$$\exists! y: \mathbb{R} \mapsto \mathbb{R}, \forall x \in \mathbb{R}, y^{\prime\prime}\left(x\right) + y^{\prime}\left(x\right) + y\left(x\right) = x,$$$$ which cannot be the case without boundary values. Thus, to specify a unique solution to your equation, you need to provide boundary values.

In the end, you conclude the following proposition as a solution to the problem: $$$$\forall y:\mathbb{R} \mapsto \mathbb{R}, \left[\forall x \in \mathbb{R}, y^{\prime\prime}\left(x\right) + y^{\prime}\left(x\right) + y\left(x\right) = x \wedge y\left(0\right) = ... \wedge y^{\prime}\left(0\right) = ...\right] \longrightarrow \left[y\left(x\right) = ...\right]$$$$

It should be noted that providing an analytical solution is not the only way to solve a differential equation (ordinary or partial). From set theory, functions are sets of 2-tuples. As long as you have a way to specify each 2-tuple in a function, the differential equation is solved. In this context, numerical approaches are also a convenient way to specify 2-tuples of a function.

A lot of times, people don't think about the logical meaning behind solving equations. Education of logic systems in math is often ignored in today's education system. Recently, someone even told me training in first order logic is not important to mathematicians. This is exactly the reason why math is so formidable to most people: using daily language to talk about a thing in the world of object language is not gonna lead to the success of most people. People fear math as math is not reachable to them. To make people truly understand the tool of math (they don't have to be masters using this tool for creative inventions), languages should be taught again from scratch.

• In regards to your last paragraph: I wish more educators thought this way. Thank you for your time! Dec 29, 2020 at 19:32
• I like your last paragraph as well :) maybe in the third displayed formula you don't really need the uniqueness? In general we will not have uniqueness of the solution. Dec 29, 2020 at 19:50
• @SeverinSchraven I agree. There is no unique solution unless there is a boundary condition. Dec 29, 2020 at 20:09
• @TaylorRendon Teaching solid logic is a task of the whole education system, instead of several educators.The educated care more about the reputation of universities than the essence of knowledge. A parent may be proud that his child goes to MIT, but he may not be proud that the child established a solid logical system for learning math. A student may be proud that he graduated from Stanford, but he may never think about the fact that the math courses he learned at Stanford are no different from those at another university. Compared to schools, true knowledge is more worth fighting for. Dec 29, 2020 at 22:45
• @Lemoine This is why I think although math is created by humans, it will be developed much better in the hands of machines, due to their uniform format of storage and endless life. The day machines can do better than us in math, we are gonna be of little use. Mar 15, 2021 at 2:58