Why are they called "first integrals"?

If I have a differential equation, $$\dot{x}(t)=f(t,x(t))$$, then the functions $$F$$ that verifies $$F(t,x(t))=const$$ are called first integral? Where does this name come from?
Also, I read here that :

The knowledge of a first integral "reduces" the number of unknowns by 1

What does this mean exactly. If I have a system with just one variable $$\dot{x}(t)=f(t,x(t))$$, and I have a first integral $$F$$,then what does it mean to reduce the number of unknowns by 1? I know that $$F(t,x(t))=const$$, but I still don't have an explicit expression for $$x(t)$$, no? If I have two variables $$x=(x_1,x_2)$$, and again I know one first integral $$F$$, why (what does it mean) to reduce de degree by one?

"When the given differential equation is of order $$n$$, and by a process of integration an equation of order $$n-1$$ involving an arbitrary constant is obtained, the latter is known as the first integral of the given equation."

Example: Consider a given differential equation $$y''(x)=f(y)$$ where $$f(y)$$ is independent of $$x$$.

This equation become integrable if you multiply $$2 y'$$ in both side of the equation.Then we get $$2 y' y'' = 2 y' f(y)$$

and after integrating we have $$(y')^{2} = 2 \int{f(y) dy} + c$$ , where c is constant of integration.

this latter equation is called the first integral of the given differential equation.

(See "Ordinary Differential Equations" by E. L. Ince )

In your case, your first integral is your solution. Actually for first order differential equation integral and solution are equivalent.

"The strength sounding name first integral is a relic of the times when the mathematicians tried to solve all differential equations by integration. In those days the name integral (or a partial integral) was given to what we now called a solution." (See "Ordinary Differential Equations" by Vladimir I. Arnol'd)

If you integrate the given differential equation for the first time you got the first integral . If you again integrate the first integral you got an equation called the second integral and so on. Ultimately by this you will get your desire solution of the differential equation.

Now the third one.

First integral reduces the order of the differential equation by one and there fore the number of unknown by one.

In particular take the previous example. Let us take $$f(y)= y^{2}$$ , then the differential equation becomes $$y''(x)= y^{2}$$. It's first integral is $$(y')^{2} = \frac {2}{3} y^{3} dy + c$$.

Suppose you are given the first integral as $$(y')^{2} = \frac {2}{3} y^{3} dy + 3$$ , then after solving this first integral you get only one unknown which you got from the last integration.

I thought by the word 'The knowledge of a first integral "reduces" the number of unknowns by 1' they mean this.

• Sorry for replying to this old post, but how did you obtain $(y')^{2} = 2 \int{f(y) dy} + c$? With respect to which variable did you integrate $2 y' y'' = 2 y' f(y)$? Seems like $y$, but since the function "$y$" doesn't depend on the variable "$y$" I don't get how you've obtained this ... But I guess I'm something anyway, since I suppose there is a reason why you've labeled both a function and a variable with the symbol $y$. May 26, 2021 at 5:02
• What do you want to mean by "the function $"y"$ doesn't depend on the variable $"y"$" ? By the way the integration is with respect to $x$. May 26, 2021 at 5:21