Introduction
In general, the laws of physics lead to the representation of phenomena through either integrals or differential equations. One has to solve them in order to obtain the corresponding description in explicit form, i.e., in order to obtain the explicit dependence on time of the variables describing the phenomenon.
Lecture 1. Technics of integration
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Integration by simple substitution (also called change of variable)
In the case the integral has the form \(\int f(g(x))g^{\prime}(x)dx\) one can change the variable \(x\) by the variable \(u=g(x)\). Thus, the differential becomes \(du=g^{\prime}(x)dx\) and the integral takes the simpler form \(\int f(u)du\). The latter integral is simpler but not necessarily easy to calculate. Remember that the result of the indefinite integral has to include the constant \(c\). Once the integral \(\int f(u)du\) has been calculated in terms of \(u\), one substitutes the expression \(u=g(x)\) to get the result in terms of the original variable \(x\). Let’s see some simple examples.
Example 1
\(\int e^{ax}dx\to\begin{cases}u=ax\ ,\ du=adx \\ \frac{1}{a}\int e^{u}du=\frac{1}{a}e^u+c \end{cases}\ ,\)
and therefore
\(\int e^{ax}dx=\frac{1}{a}e^{ax}+c\ .\)
Example 2
\(\int \sin(ax)dx\to\begin{cases}u=ax\ ,\ du=adx \\ \frac{1}{a}\int \sin(u)du=-\frac{1}{a}\cos(u)+c \end{cases}\ ,\)
and therefore
\(\int \sin(ax)dx = -\frac{1}{a}\cos(ax)+c \ .\)
Example 3
\(\int e^{ax^2}xdx\to\begin{cases}u=ax^2\ ,\ du=2axdx \\ \frac{1}{2a}\int e^{u}du=\frac{1}{2a}e^u+c \end{cases}\ ,\)
and therefore
\(\int e^{ax^2}xdx=\frac{1}{2a}e^{ax^2}+c\ .\)
Example 4
\(\int \frac{\sin(a/x)}{x^2}dx\to\begin{cases}u=a/x\ ,\ du=-\frac{a}{x^2}dx \\ -\frac{1}{a}\int \sin(u)du=\frac{1}{a}\cos(u)+c \end{cases}\ ,\)
and therefore
\(\int \frac{\sin(a/x)}{x^2}dx = \frac{1}{a}\cos(a/x)+c \ .\)
Example 5
\(\int \sqrt{a^2+x^2}\ xdx\to\begin{cases}u=a^2+x^2\ ,\ du=2xdx \\ \frac{1}{2}\int \sqrt{u}\ du=\frac{1}{2}\frac{2}{3}u^{3/2}+c \end{cases}\ ,\)
and therefore
\(\int \sqrt{a^2+x^2}\ xdx\ = \frac{1}{3}\left(a^2+x^2\right)^{3/2}+c \ .\)
Integration by parts
This method is based on the rule for the differentiation of the product of two functions \(d\left(fg\right)=f^{\prime}gdx+fg^{\prime}dx\). Integration on both sides leads to \(fg=\int f^{\prime}gdx+\int fg^{\prime}dx\). If one of the integrals on the right side is easy to calculate, then the other one can be obtained directly without further integration. One has to decide what to take as \(f\) and what as \(g^{\prime}\). This requires some experience. Here there are some examples:
Example 1
\(\int xe^{ax}dx\to\begin{cases}f=x\Rightarrow f^{\prime}=1\\ g^{\prime}=e^{ax}\Rightarrow g=\frac{1}{a}e^{ax}\end{cases}\)
\(\Rightarrow x\frac{1}{a}e^{ax}=\int 1*e^{ax}dx+\int xe^{ax}dx\)
\(\Rightarrow \int xe^{ax}dx=\frac{1}{a}xe^{ax}-\int e^{ax}dx\)
\(\Rightarrow \int xe^{ax}dx=\frac{1}{a}xe^{ax}-\frac{1}{a}e^{ax}+c\)
Let’s see what would happen if we make a different selection:
\(\int xe^{ax}dx\to\begin{cases}f=e^{ax}\Rightarrow f^{\prime}=ae^{ax}\\ g^{\prime}=x\Rightarrow g=\frac{1}{2}x^2\end{cases}\)
\(\Rightarrow e^{ax}\frac{1}{2}x^2=\int ae^{ax}*\frac{1}{2}x^2dx+\int e^{ax}xdx\)
\(\Rightarrow \int xe^{ax}dx=\frac{1}{2}x^2e^{ax}-\frac{a}{2}\int x^2e^{ax}dx\)
We see that the integral in the right side is more difficult to calculate than the original one. There is nothing wrong, but the transformation doesn’t help in calculating the original integral.
Lecture 2. Ordinary differential equations
homogeneous differential equations