📂Odinary Differential Equations

Second-Order Linear Homogeneous Differential Equations with Constant Coefficients and Characteristic Equation

Theorem¹

The general solution of a second-order linear homogeneous differential equation with constant coefficients $a y^{\prime \prime} + by^\prime +cy=0$ is as follows.

$$y(x)=A e^{r_{1} x}+Be^{r_2 x}$$

At this time, $r_{1,2}=\dfrac{-b \pm \sqrt{b^2-4ac}} {2a}$

Corollary

The solution of $a y^{\prime \prime} + cy = 0$ is as follows.

$$y(x) = A e^{i\sqrt{\frac{c}{a}} x}+Be^{-i\sqrt{\frac{c}{a}} x} = C\cos{\textstyle (\sqrt{\frac{c}{a}}x)} + D\sin{\textstyle (\sqrt{\frac{c}{a}}x)}$$

Solution

$$\begin{equation} a\dfrac{d^2}{dx^2}y+b\dfrac{d}{dx}y+cy = 0 \label{eq1} \end{equation}$$

First, let’s define the differential operator $D$ as follows.

$$D:=\dfrac{d}{dx} \\ Df = D(f) = \dfrac{df}{dx}$$

Then, since $D$ satisfies $D(ay_{1}+y_{2}) = a\dfrac{dy_{1}}{dx} + \dfrac{dy_{2}}{dx} = aDy_{1}+Dy_{2}$, it is a linear operator. Using $D$ to express Equation $\eqref{eq1}$ gives the following.

$$\begin{align} &&aD^2y+bDy+cy&=0 \\ \implies&& (aD^2+bD+c)y&=0 \end{align}$$

If there is a constant $r$ that satisfies $Dy=ry$, then we obtain the following equation from the above equation.

$$(aD^2+bD+c) y = (ar^2+br+c) y = 0$$

Since we are looking for solutions that satisfy $y \ne 0$, we obtain the following condition

$$aD^{2} + bD + c = ar^{2}+br+c = 0$$

This quadratic equation is called the characteristic equation.

$$\begin{align*} r_{1} &= \dfrac{-b + \sqrt{b^2-4ac}} {2a} \\ r_2 &=\dfrac{-b - \sqrt{b^2-4ac}} {2a} \end{align*}$$

Let’s say $r_{1}, r_{2}$ are two different real numbers. Then, from the above equations, we obtain the following.

$$(aD^2 + bD+c)y=0 \implies \ a(D-r_{1})(D-r_2)y=0$$

• Case 1. $(D-r_{1})y=0$

  If $\dfrac{dy}{dx}=r_{1}y$ and we find $y$ through the method of separation of variables, then

  $$y_{1}(x)=Ae^{r_{1}t}$$

• Case 2. $(D-r_2)y=0$

  Similarly, if we find $y$, then

  $$y_{2}(x)=Be^{r_2t}$$

If $y_{1}$ and $y_{2}$ are solutions to the given differential equation, then $y_{1}+y_{2}$ is also a solution, so the general solution of the given differential equation is

$$y(x)=y_{1}+y_{2}=Ae^{r_{1}t} + Be^{r_2t}$$

■