First-order differential equations

We discuss some facts about first-order differential equations for beginners. Such equations are important because many problems in our real life can be modeled as a differential equation.

We assume that the reader is familiar with the concept of continuous function primitives.

First-order differential equations with constant coefficients

In algebra, we already studied algebraic equations where a variable is a number. Here we study equations where the variable is a function.

Let $a$ be a real number and $f:\mathbb{R}\to \mathbb{R}$ be a continuous function. We look for differential functions $u:\mathbb{R}\to \mathbb{R}$ such that$$\begin{array}{r}\dot{u}(x)=au(x)+f(x).\end{array}$$ Here we denote $\dot{u}(x)=\frac{d}{dx}u(x)$, the derivative of the function $u$.

We recall that $\frac{d}{dx}{e}^{-ax}=a{e}^{-ax}.$ By multiplying the both sides of the above differential equation by $e^{-ax}$, we obtain $e^{-ax}\dot{u}(x)-a{e}^{-ax}u(x)=e^{-ax}f(x)$. Also, we write$$\begin{array}{r}\frac{d}{dx}(e^{-ax}u(x))=e^{-ax}f(x).\end{array}$$By taking the integral between $0$ and $x$ in the both sides of this equation, we get $$\begin{array}{r}{e}^{-ax}u(x)=u(0)+{\int }_0^x{e}^{-as}f(s)ds.\end{array}$$ Now by multiplying the both sides of the avove equality br $e^{ax},$ we obtain $$\begin{array}{r}u(x)=e^{ax}u(0)+{\int }_0^x{e}^{a(x-s)}f(s)ds.\end{array}$$

Cauchy problem: A first-order differential is a Cauchy problem if it take the following form $$\begin{array}{r}\{\begin{array}{ll}\dot{u}(t)=au(t),& t\in \mathbb{R},\\ u(t_0)=x.\end{array}\end{array}$$ Here $t_0$ is the initial time and $x$ is the initial state. The solution to this Cauchy problem is $u(t)=e^{ta}x$.

Equations with variable coefficients

In most cases the coefficients of differential equations are functions. These equations take the following form $$\begin{array}{r}a(x)\dot{u}(x)+b(x)u(x)=0,\end{array}$$where $a(\cdot )$ and $b(\cdot )$ are continuous function such that $a(x)\ne 0$. This equation can be rewritten as$$\begin{array}{r}\frac{\dot{u}(x)}{u(x)}=-\frac{b(x)}{a(x)}.\end{array}$$ On the other hand, we recall that $$\begin{array}{r}\frac{d}{dx}\ln (|u(x)|)=\frac{\dot{u}(x)}{u(x)}.\end{array}$$ We then obtain $$\begin{array}{r}\ln (|u(x)|)=\int -\frac{b(x)}{a(x)}dx.\end{array}$$ Hence, we have$$\begin{array}{r}u(x)=e^{-{\displaystyle \int \frac{b(x)}{a(x)}dx}}.\end{array}$$

Let consider the first example $(x-1)u^{\prime }-2u=0$, here we have $a(x)=x-1$ and $b(x)=-2$. Then the solution is given by $$\begin{array}{r}u(x)=e^{{\displaystyle \int \frac{2}{x-1}dx}}=e^{(2\ln (|x-1|+C)}=A{e}^{\ln ((x-1{)}^2)}.\end{array}$$ Thus the solution of the differential equation is $u(x)=A(x-1{)}^2$, where $A$ is a real constante.