Homogeneous linear differential equations with constant coefficients

Consider the homogeneous linear differential equation

$$\begin{array}{}\text{(Eq:code)}& y^{(n)}+a_{n-1}{y}^{(n-1)}+\cdots +a_1{y}^{\prime }+a_{0}y=0\end{array}$$

where $a_0,a_1,\cdots ,a_{n-1}\in \mathbb{R}$ are constants. 

Using the differential operator $D=\frac{d}{dx}$ and its polynomial

$$\begin{array}{}\text{(Eq:epoly)}& E=D^n+a_{n-1}{D}^{n-1}+\cdots +a_{1}D+a_0,\end{array}$$

(Eq:code) can be expressed as

$$Ey=0.$$

Now, consider the polynomial of variable $t$

$$F(t)=t^n+a_{n-1}{t}^{n-1}+\cdots +a_{1}t+a_0.$$

Then, we have

$$E=F(D)$$

and (Eq:code) is expressed as

$$F(D)y=0.$$

We can use the properties of polynomials to solve this type of differential equation. We need some results from algebra.

Definition (Relatively prime, coprime)

Polynomials $F_1(t)$ and $F_2(t)$ are said to be relatively prime or coprime if the equations $F_1(t)=0$ and $F_2(t)=0$ do not have common solutions within $\mathbb{C}$.

Lemma

The polynomials $F_1(t)$ and $F_2(t)$ in $t$ are relatively prime if and only if there exist polynomials $G_1(t)$ and $G_2(t)$ such that

$$G_1(t)F_1(t)+G_2(t)F_2(t)=1.$$

Proof. Omitted. ■

Remark. The proof of this lemma is based on the Euclidean Algorithm for finding the greatest common divisor (GCD) of polynomials. The right-hand side of the above equation being equal to 1 indicates that the GCD of $F_1(t)$ and $F_2(t)$ is 1 (or any constant, but not a polynomial of $t$), that is, they are relatively prime (just like integers). □

See also: Polynomial greatest common divisor (Wikipedia)

Example. Let $F_1(t)=t-1$ and $F_2(t)=t-2$. These are clearly relatively prime, and

$$G_1(t)(t-1)+G_2(t)(t-2)=1$$

where $G_1(t)=1$ and $G_2(t)=-1$. □

Example. Let $F_1(t)=t^3-3t^2+2t+3$ and $F_2(t)=t^2+1$. By dividing $F_1(t)$ by $F_2(t)$, we have

$$\begin{array}{}\text{(eg. 1)}& F_1(t)=(t-3)F_2(t)+t+6.\end{array}$$

By dividing $F_2(t)=t^2+1$ by $t+6$, we have

$$\begin{array}{}\text{(eg. 2)}& F_2(t)=(t-6)(t+6)+37.\end{array}$$

From this, we can see that the greatest common divisor of $F_1(t)$ and $F_2(t)$ is 1 (we use the convention that the coefficient of the leading term of a GCD is 1), which means they are relatively prime.

From (eg. 2), 

$$37=F_2(t)-(t-6)(t+6).$$

Using (eg. 1),

$$\begin{array}{rcl}37& =& F_2(t)-(t-6)(F_1(t)-(t-3)F_2(t))\\ & =& -(t-6)F_1(t)+(t^2-9t+19)F_2(t).\end{array}$$

Setting $G_1(t)=-\frac{t-6}{37}$ and $G_2(t)=\frac{t^2-9t+19}{37}$, we have

$$G_1(t)F_1(t)+G_2(t)F_2(t)=1.$$

□

Example. Let $F_1(t)=t^3-8=(t-2)(t^2+2t+4)$ and $F_2(t)=t^2-4t+4=(t-2{)}^2$. The greatest common divisor is $t-2$, and hence they are not relatively prime. Suppose there are polynomials $G_1(t)$ and $G_2(t)$ such that $$G_1(t)F_1(t)+G_2(t)F_2(t)=1$$ holds. The left-hand side is divisible by $t-2$, whereas the right-hand side is not. This is a contradiction. Therefore, there are no such polynomials as $G_1(t)$ and $G_2(t)$. □

Theorem (Solution of $F_1(D)F_2(D)y=0$ when $F_1(t)$ and $F_2(t)$ are relatively prime)

Let $F_1(t)$ and $F_2(t)$ be polynomials in $t$ that are relatively prime, and $F(t)=F_1(t)F_2(t)$. Then, any solution of $F(D)y=0$ is of the form $y=y_1+y_2$ where $y_1$ and $y_2$ are solutions of $F_1(D)y=0$ and $F_2(D)y=0$, respectively. Conversely, any function of the form $y=y_1+y_2$, where $y_1$ and $y_2$ are solutions of $F_1(D)y=0$ and $F_2(D)y=0$, respectively, is a solution of $F(D)y=0$.

Proof. Suppose $y$ is a solution of $F(D)y=F_1(D)F_2(D)y=0$. By the above Lemma, there exist polynomials $G_1(t)$ and $G_2(t)$ such that $G_1(t)F_1(t)+G_2(t)F_2(t)=1$. Using these polynomials, let us define

$$\begin{array}{rcl}{y}_1& =& G_2(D)F_2(D)y,\\ y_2& =& G_1(D)F_1(D)y.\end{array}$$

Then, we have $y=y_1+y_2$. Now,

$$\begin{array}{rcl}{F}_1(D)y_1& =& F_1(D)G_2(D)F_2(D)y\\ & =& G_2(D)F_1(D)F_2(D)y\text{(commutative law)}\\ & =& G_2(D)F(D)y=0.\end{array}$$

Thus, $y_1$ is a solution of $F_1(D)y=0$. Similarly, $y_2$ is a solution of $F_2(D)y=0$.

Conversely, suppose $y_1$ and $y_2$ are solutions of $F_1(D)y=0$ and $F_2(D)y=0$, respectively, and let $y=y_1+y_2$. Then,

$$\begin{array}{rcl}F(D)y& =& F(D)(y_1+y_2)\\ & =& F_2(D)F_1(D)y_1+F_1(D)F_2(D)y_2\\ & =& 0.\end{array}$$

Thus, $y$ is a solution of $F(D)y=0$. ■

Example. Let us find the general solution of

$$y^{″}-3y^{\prime }+2y=0.$$

Using the differential operator, this can be written as

$$(D^2-3D+2)y=0.$$

The operator can be factorized as

$$D^2-3D+2=(D-1)(D-2)$$

and $D-1$ and $D-2$ are relatively prime. The solution of $(D-1)y=0$ is $y=C_1{e}^x$, and that of $(D-2)y=0$ is $y=C_2{e}^{2x}$. Thus, the general solution of $(D-1)(D-2)y=0$ is

$$y=C_1{e}^x+C_2{e}^{2x}$$

where $C_1$ and $C_2$ are constants. □

Complex-valued functions

To treat more general linear equations, it is convenient to use complex-valued functions. Before proceeding, please read the Calculus of complex-valued functions.

We use the following theorem without proof.

Theorem (Fundamental Theorem of Algebra)

A polynomial equation of degree $n$ with coefficients in $\mathbb{R}$,

$$a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0=0,a_0,a_1,\cdots ,a_n\in \mathbb{R},a_n\ne 0,$$

has exactly $n$ (possibly redundant) solutions in $\mathbb{C}$. 

This theorem implies that it is always possible to factorize the above polynomial of degree $n$ as

$$a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0=a_n(x-{\alpha }_1{)}^{m_1}(x-{\alpha }_2{)}^{m_2}\cdots (x-{\alpha }_l{)}^{m_l}$$

where $l$ is some natural number, ${\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_l\in \mathbb{C}$ are the distinct solutions, and $m_1,m_2,\cdots ,m_l\in \mathbb{N}$ are the multiplicities of the solutions with $m_1+m_2+\cdots +m_l=n$.

Remark. It is easy to show that if $\alpha \in \mathbb{C}$ is a solution of a polynomial equation with real coefficients, then its complex conjugate $\overline{\alpha }$ is also a solution (exercise!) □

Theorem (Solution of $(D-\alpha {)}^{m}y=0$)

Let $F(t)=(t-\alpha {)}^m$ where $\alpha \in \mathbb{C}$ and $m\in \mathbb{N}$. The general solution of the differential equation $F(D)y=0$ is given by

$$\begin{array}{}\text{(eq:sol1)}& y=c_0{e}^{\alpha x}+c_{1}x{e}^{\alpha x}+\cdots +c_{m-1}{x}^{m-1}{e}^{\alpha x}\end{array}$$

where $c_0,c_1,\cdots ,c_{m-1}\in \mathbb{C}$ are constants.

Proof. Let $h(x)$ be an arbitrary function of $x$. We have

$$(D-\alpha )h(x)e^{\alpha x}=h^{\prime }(x)e^{\alpha x}+h(x)\alpha e^{\alpha x}-\alpha h(x)e^{\alpha x}=h^{\prime }(x)e^{\alpha x}.$$

By repeating this, we have, in general

$$(D-\alpha {)}^{k}h(x)e^{\alpha x}=h^{(k)}(x)e^{\alpha x}.$$

If $h(x)$ is a polynomial of degree less than $m$, then $(D-\alpha {)}^{m}h(x)e^{\alpha x}=0$. Therefore, the function of the form of (eq:sol1) is a solution of $F(D)y=0$.

Conversely, suppose that $y=y(x)$ is a solution of $F(D)y=0$ and let $h(x)=y(x)e^{-\alpha x}$. We have $y(x)=h(x)e^{\alpha x}$. From what we have shown above,  $F(D)y=h^{(m)}(x)e^{\alpha x}$. But $F(D)y=0$ so $h^{(m)}(x)e^{\alpha x}=0$, and hence $h^{(m)}(x)=0$. This indicates that $h(x)$ is a polynomial of degree at most $(m-1)$. Thus, $y(x)=h(x)e^{\alpha x}$ has the form of (eq:sol1). ■

Example (Harmonic oscillator). The equation of motion of a harmonic oscillator with spring constant $k$ and mass $m$ is given by

$$m\frac{d^{2}x}{d{t}^2}=-kx.$$

Let $\omega =\sqrt{\frac{k}{m}}$. This become

$$(D^2+{\omega }^2)x=0.$$

But $D^2+{\omega }^2=(D-i\omega )(D+i\omega )$, so we need to solve 

$$(D-i\omega )x=0$$

and

$$(D+i\omega )x=0.$$

From the former, we have $x_1(t)=C_1{e}^{i\omega t}$. From the latter, we have $x_2(t)=C_2{e}^{-i\omega t}$. Therefore, the general solution is 

$$x(t)=C_1{e}^{i\omega t}+C_2{e}^{-i\omega t}.$$

But, we note that $x(t)$ should be a real function because it describes a physical quantity (the coordinate of the mass point). Let's rewrite the above solution using Euler's formula,

$$x(t)=(C_1+C_2)\cos (\omega t)+i(C_1-C_2)\sin (\omega t).$$

For this to be a real function, $C_1+C_2$ must be real and $C_1-C_2$ must be purely imaginary. This can be achieved if we set $C_2=\overline{C_1}$ (i.e., $C_1$ and $C_2$ are complex conjugate of each other). By setting $A=C_1+C_2$ and $B=i(C_1-C_2)$, we can rewrite the solution as

$$x(t)=A\cos \left(\sqrt{\frac{k}{m}}t\right)+B\sin \left(\sqrt{\frac{k}{m}}t\right).$$

This is indeed a real-valued function if $A$ and $B$ are real constants. □

Example. Let us solve

$$y^{‴}+y^{″}+y^{\prime }-3y=0.$$

Let $F(t)=t^3+t^2+t-3$. The above ODE is $F(D)y=0$. Since 

$$F(t)=(t-1)(t^2+2t+3)=(t-1)(t-\alpha )(t-\overline{\alpha })$$ where $\alpha =-1+i\sqrt{2}$, the general solution is the sum of the solutions of $(D-1)y=0$, $(D-\alpha )y=0$, and $(D-\overline{\alpha })y=0$. From these, we have $y=C{e}^x$, $y=C_1{e}^{\alpha x}$, and $y=C_2{e}^{\overline{\alpha }x}$, respectively. Thus, the general solution is

$$y=C{e}^x+C_1{e}^{(-1+i\sqrt{2})x}+C_2{e}^{(-1-i\sqrt{2})x}.$$

To make this a real function, we set $A=C_1+C_2$ and $B=i(C_1-C_2)$ and use Euler's formula to have

$$y=C{e}^x+A{e}^{-x}\cos (\sqrt{2}x)+B{e}^{-x}\sin (\sqrt{2}x).$$

Note that this is a real-valued function if all the constants $A,B$, and $C$ are real. □