How is $y'=x^2+1$ a differential equation? From James Stewart Essential Calculus Early Transcendentals Textbook,

A differential equation is an equation that contains an unknown function and one or more of its derivatives.

But the equation
$y'=x^2+1$
doesn't contain the unknown function $y$, it only contains $y'$ and $x$. And yet my textbook says its a DE of order 1.
Even Wikipedia says a DE "relates one or more functions and their derivatives", but $y'=x^2+1$ doesn't contain the function $y$ so how is it a DE ?
 A: That definition is not a very good one, for the reason you're pointing out here. To be precise, an ordinary differential equation of positive integer order $n$ is one which can be written in the form
$$ F\left(x, y, y',\dots,y^{(n)}\right) = 0 $$
where $y$ is an unknown function of the single variable $x$, and $\frac{\partial F}{\partial y^{(n)}} \not\equiv 0$. The upshot of that condition is that the degree will be equal to the highest order of derivative present in the equation, barring cancellations.

For your example given,
$$ F\left(x,y, y'\right) = y' - x^2 - 1 $$
with $\frac{\partial F}{\partial y'} = 1 \neq 0$, so it is a first order ordinary differential equation. This should align with your instinct upon inspection, even if it technically disagrees with Stewart's definition.
A: I think, the following definition may clear your doubts in all matter. This is taken from "An Introduction to Differential Equations and Their Applications" by Stanley J Farlow.


*

*A differential equation is an equation that relates the derivatives of an unknown function, the function itself, the variables by which the function is defined, and constants.

*If the unknown function depends on a single real variable, the differential equation is called an ordinary differential equation.

*A partial differential equation is one that contains partial derivatives with respect to more than one independent variable.


An $n^\text{th}$-order ordinary differential equation is an equation that has the
general form $$F\left(x, y, y',\cdots,y^{(n)}\right) = 0\tag1$$where the unknown quantity $y = y(x)$ is called the dependent variable, the real variable, $x$, is called the independent variable, the primes $"~'~"$ denote differentiation with respect to $x$, that is, $y' = \frac{∂y}{∂x},~~ y'' = \frac{d^2y}{dx^2}$ and so on.
Now lets come to your equation, which can be written as $~y'=x^2+1\implies y'-x^2-1=0~.$ So here it contains derivatives of an unknown function $~y~$, which is $y'$, the variable $~x,~$ and the constant $1~.$ Hence by the definition given above it is a differential equation and in particular it is an ordinary differential equation.
