Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

1: ODE Basics

The goal of this chapter is to extend the systematic approach of integration to include other types of mathematical functions involving relative change of variables. The process of identifying which dimension is changing with respect to another one is the first step in the systematic approach. The second step involves defining an instantaneous change for the function by writing what is called a differential. Differentials are of the form:

\[dF\,=\,f(x) \cdot dx.\]

Dividing both sides by the infinitely small, \(dx\), defines the derivative or instantaneous rate of change of \( F(x)\) with respect to \(x\). What is important to understand is that \( f(x)\) is assumed to be constant over the interval, \(dx\). Thus the infinite sum of these \(dx\), \(s\) is given by integrating the function. The net change in the function, \(F(x)\) , over an interval form \(x = a\) to \(x = b\) is thus:

\[\int_a^b\,F(x)=\sum_{i=1}^{\infty}f(x_i)\,dx = \int^b_a f(x)\,dx.\]

Let us now look at various combinations of \(f(x)\) and \(x\), and see how we differentials are defined for each case. Consider case 1, where some situation, \(F\) is a function of two independent conditions \(f\) and \(x\),

\[F\,=\, f \cdot x.\]

If \(f\) is a function of \(x\), \(f(x)\), then its value changes as \(x\) changes. The differential is defined by converting the independent \(x\) into an infinitely small \(dx\).

\[dF\,=\,f(x) \cdot dx\]

The differential is then integrated to determine how much the situation, \(F\),  changes as \(x\) changes from \(x = a\) to \(x = b\).

Consider case 2 where \(F\) is a function of two independent conditions, \(f\) and \({x}^n\)

\[F\,=\,f \cdot x^n.\]

In this case, \(x^n\) represents a fixed or constant condition. We can replace \(A\) with \(x^n\) to get:

\[F(f\,,\,A)\,=\,f \cdot A\]

If \(f\) is given as a function of \(x\), then we will first have to re-write \(f(x)\) as a function of \(A\). We know that:


\[\text{therefore } x\,=\,\sqrt[n]{A}.\]

If \(f(x)=x^m\), then \(f(A)=\left(\sqrt[n]{A}\right)^m\). Once we have \(f(A)\), we can write our differential;

\[dF\,=\,f(A) \cdot dA.\]

We can then integrate this function from some value of \(A_1\) to \(A_2\). Keep in mind that, \(A_1=x_1^n\)  \(A_2=x_2^n\)

\[\int_{A_1}^{A_2}F(A)\,=\,\int_{A_1}^{A_2}f(A) \cdot dA.\]

This approach to writing and integrating the differential for case 2 functions is confusing and abstract. An easier and more logical approach is to integrate with respect to \(x\) only: For example if,

\[F\,=\,f \cdot x^n.\]

First let \(A=x^n\), then \(\frac{dA}{dx}=n \cdot x^{n-1}\) or:


Substituting this back into the function yields:

\[dF\,=\,f \cdot dA\]

\[dF\,=\,f \cdot nx^{n-1}dx.\]

If \(f\) is a constant then from \(x =a\) to \(x = b\), the change in \(F\) is:

\[\Delta F\,=\,\int_a^bf \cdot nx^{n-1}dx\,=\,\int_a^bfx^n.\]

The result is the same function we started with. If \(f\) were a function of \(x\), then its value would change as \(x\) goes from \(a\) to \(b\). The differential can be written as:


Integrating this function over an interval from \(x = a\) to \(x = b\) results in

\[\int_a^b F(x)\,=\,\int_a^bf\left(x\right).nx^{n-1}dx.\]

This results in the same numerical answer as the previously defined method of writing the differential.

To summarize, if \(F=f \cdot x^n\)  ; and \(f\) is a function of \(x\), then:

\[dF\,=\,f\left(x\right) \cdot d(x^n)\]

\[\Delta F\,=\,\int_a^bf\left(x\right) \cdot nx^{n-1}dx.\]

Here \(A = x^n\), where \(dA=nx^{n-1} \cdot dx\)

The third case of writing a differential is for functions of the form:

\[F\,=\,f\left(x\right) \cdot x^{-n}\]


Once again, \(\frac{1}{x^n}\) is a constant factor that defines \(F\). We can let \(A(x)=\frac{1}{x^n}\)  , then:


\[dA\,=\,-nx^{-n-1} \cdot dx.\]

Substituting this back into case three:

\[\int_a^b F(x)\,=\,\int_a^bf\left(x\right) \cdot -nx^{-n-1}dx.\]

Our results can be generalized as follows. If \(F\) is a function of two variables, \(f\) and \(x\) such as:

\[F\left(f\,,\,A\right)\,=\,f \cdot A.\]

If \(A\) is some function of \(x\), and \(f\) is also a function of \(x\) then both these values change as \(x\) changes. It is important to understand that \(A(x)\) is a constant value, dependent on \(x\), that defines \(F\). Therefore, the differential is defined as:

\[dF\,=\,f(x) \cdot dA.\]

Since \(A\) is a function of \(x\), then:



Substituting this back into the differential gives us the general expression:

\[dF\,=\,f(x) \cdot A'(x) \cdot dx.\]

We can integrate this over any interval of \(\Delta x\) to find the net change in \(F\) over the interval. Our results can be generalized for any combination of \(f(x)\) and \(A(x)\), where \(f(x)\) and \(A(x)\) can represent any function of \(x\). The differential and integral is:

\[F\,=\,f(x) \cdot A(x)\]

\[dF\,=\,f(x) \cdot A'(x) \cdot dx.\]