Higher order homogeneous linear differential equation

Note that $y = ce^{mx}$ is a complete solution of $y^{\prime} = my$. Then we use $y = e^{mx}$ as a candidate for the solution of the following $n$th-order linear differential equation.

$$L(y) = a_{n}y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_{1}y^{\prime} + a_{0}y = 0$$

Then since

$$L(e^{mx}) = a_{n}m^{n}e^{mx} + a_{n-1}m^{n-1}e^{mx} + \cdots + a_{1}me^{mx} + a_{0}e^{mx} = 0$$

$e^{mx}$ is a solution of $L(y) = 0$ if and only if

$$P(m) = a_{n}m^{n} + a_{n-1}m^{n-1} + \cdots + a_{1}m + a_{0} = 0$$

The polynomial $P(m)$ is called the characteristic polynomial, and $P(m) = 0$ is called the characteristic equation. This way, we only need to solve the polynomial equation $P(m) = 0$ instead of solving $L(y) =$.

Example 2..7   Solve the differential equation $y^{\prime\prime} + 3y^{\prime} + 2y = 0$.

SOLUTION The roots of the characteristic equation $m^{2} + 3m + 2 = 0$ are $m = -1, m = -2$. Then $e^{-x}$ and $e^{-2x}$ are solutions and by example 2.2, these solutions are linearly independent. Thus, the general solution is

$$y = c_{1}e^{-x} + c_{2}e^{-2x}\ensuremath{\ \blacksquare}$$

Theorem 2..8   Suppose that the roots of the characteristic polynomial $P(m) = 0$ of the $n$th -order linear differential equation are distinct real roots $m = r_1, r_2,\ldots, r_n$. Then the set of solutions $\{e^{r_{1}x}, e^{r_{2}x}, \cdots, e^{r_{n}x}\}$ is a basis for the solution space.

Proof. We show $\{e^{r_{1}x}, e^{r_{2}x}, \cdots, e^{r_{n}x}\}$ are linearly independent by using Wronskian. Then

$$W = \begin{array}{\vert llll\vert} e^{r_{1}x} & e^{r_{2}x} & \cdo... ...s & \vdots & \vdots\\ r_{1}^{n-1} & r_{2}^{n-1} & \cdots & r_{n}^2 \end{array}$$

Now this determinant is the Vandermonde determinant. Thus we have $W =\Pi_{1 \leq i,j \leq n-1}(r_j - r_i)$. Since $r_i$ are distinct real roots, $W \neq 0$.

Example 2..8   Suppose the roots of the characteristic equation $P(m) = 0$ of the 4th-order homogeneous linear differential equation $L(y) = 0$ are $m = 1,2,3,4$. Then find the general solution.

SOLUTION By the theorem above, $\{e^x, e^{2x}, e^{3x}, e^{4x}\}$ is the basis of the solution space. Thus, the general solution is the linear combination of $\{e^x, e^{2x}, e^{3x}, e^{4x}\}$. Therefore,

$$y = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} + c_4 e^{4x}\ensuremath{\ \blacksquare}$$

Example 2..9   Solve the differential equation $y^{\prime\prime} -2y^{\prime} + y = 0$.

SOLUTION The roots of the characteristic equation $m^{2} -2m + 1 = 0$ are $m = 1,1$. Then $y_{1} = e^{x}$ is a solution. Since this differential equation is the 2nd-order, we must have another linearly independent solution. By exercise 2.2.1, $y_{2}$ can be obtained by

$$y_{2} = e^{x}\int\frac{e^{-\int-2dx}}{e^{2x}} dx$$

$$= e^{x}\int\frac{e^{2x}}{e^{2x}} dx = xe^{x}$$

Thus the general solution is

$$y = c_{1}y_{1} + c_{2}y_{2} = c_{1}e^{x} + c_{2}xe^{x}\ensuremath{\ \blacksquare}$$

Theorem 2..9   Suppose the roots of the characteristic equation $P(m) = 0$ are $k$-fold multiple roots $m = r$. Then $x^{n}e^{rx}, \ (n = 0,1,\ldots,k-1)$ are the solutions of the differential equation $L(y) = 0$. Furthermore, the set of solutions $\{x^{n}e^{rx} : n = 0,1,\ldots,k-1)\}$ is a basis of the solution space.

Proof Denote $D = \frac{d}{dx}$. Then we can express

$$L(y) = a_{n}y^{(n)} + a_{n-1}y^{(n-1)} + \cdots + a_{1}y^{\prime} + a_{0}y$$

as

$$L(D)y = (a_{n}D^{n} + a_{n-1}D^{n-1} + \cdots + a_{1}D + a_{0})y$$

Since $L(e^{mx}) = e^{mx}P(m)$, we have $L(D)e^{mx} = e^{mx}P(m)$. Now evaluate $D^{i}(x^{n}e^{mx})$. Then

$$D^{i}(x^{n}e^{mx}) = e^{mx}(D + m)^{i}x^{n}$$

Then

$$L(D)x^{n}e^{mx} = e^{mx}P(D + m)x^{n}.$$

Thus for $m = r$ is $k$-fold multiple roots, $L(D) = (D - r)^{k}$ and

$$L(D)x^{n}e^{rx} = e^{rx}D^{k}x^{n} = 0 \ (n = 1,2,\ldots,k-1). \$$

Next we show $x^{n}e^{rx} \ (n = 0,1,\ldots,k-1)$ are linearly independent.

$$c_{1}e^{rx}+c_{2}xe^{rx} + \cdots + c_{k-1}x^{k-1}e^{rx} = 0$$

Since $r^{rx} \neq 0$, we have

$$c_{1} + c_{2}x + c_{3}x^{2} + \cdots + c_{k-1}x^{k-1} = 0.$$

Thus, $c_{1} = c_{2} = c_{3} = \cdots = c_{k-1} = 0$. $\ \blacksquare$

Example 2..10   If the roots of the characteristic equation $P(m) = 0$ of the 4th-order linear differential equation $L(y) = 0$ are $m = -3,-3,-3,-3$, then find the general solution.

SOLUTION By the theorem 2.9, $e^{-3x},xe^{-3x},x^{2}e^{-3x},x^{3}e^{-3x}$ are linearly independent solutions of $L(y) = 0$. Thus the general solution is

$$y = c_{1}e^{-3x} + c_{2}xe^{-3x} + c_{3}x^{2}e^{-3x} + c_{4}x^{3}e^{-3x}\ensuremath{\ \blacksquare}$$

Let the coefficients of $L(y)$ be real. If $a + bi$ is the root of the characteristic equation $P(m) = 0$, then the conjugate $a - bi$ is also a root. Thus,

$$y_{1} = e^{(a+bi)x}, y_{2} = e^{(a-bi)x}$$

are solutions of $L(y) = 0$. Now by the Euler's formula, $e^{i\theta} = \cos{\theta} + i\sin{\theta}$, and the linearlity of solutions

$$y_{3} = \frac{y_{1} + y_{2}}{2} = e^{ax}\cos{bx},$$

$$y_{4} = \frac{y_{1} - y_{2}}{2i} = e^{ax}\sin{bx}$$

are solutions of $L(y) = 0$. It is not hard to show $\{y_{3}$ and $y_{4}\}$ are linearly independent. Thus, the basis of the solutions corresponding to $k$-fold multiple roots of complex numbers is

$$\{e^{ax}\cos{bx},e^{ax}\sin{bx},xe^{ax}\cos{bx},x^{ax}\sin{bx},\ldots,x^{k-1}e^{ax}\cos{bx},x^{k-1}e^{ax}\sin{bx}\}.$$

Example 2..11   Solve the differential equation $y^{\prime\prime} + 2y^{\prime} + 4y = 0$.

SOLUTION The roots of the characteristic equation $m^{2} + 2m + 4 = 0$ are $m = -1\pm i \sqrt{3}$. Then

$$\{e^{-x}\cos{\sqrt{3}x},e^{-x}\sin{\sqrt{3}x}\}$$

is the fundamental solution. Thus the general solution is

$$y = c_{1}e^{-x}\cos{\sqrt{3}x} + c_{2}e^{-x}\sin{\sqrt{3}x}\ensuremath{\ \blacksquare}$$

The mass of the object is $m$, the spring constant is $k$, The force loss due to friction of dashpot is proportional to the speed of the object $dy/dt$. Then the forces acting on the object is given by

Figure 2.2: vibration of spring

1. $F_{1}$	= $mg$ the gravitational force
2. $F_{2}$	= $-ky$ the restoring force by the spring
3. $F_{3}$	= $- a \frac{dy}{dt} \ (a > 0)$ force due to friction
4. $F_{4}$	= $F(t)$ the external force

Now using the Newton's 2nd law, we have

$$m\frac{d^2 y}{dt^2} = mg - ky - a \frac{dy}{dt} + F(t).$$

or

$$m\frac{d^2 y}{dt^2} + a \frac{dy}{dt} + ky = F(t) + mg$$

Now note that when the external force $F(t) = 0$, $ky_0 = mg$. Then for $y + y_0$, we have

$$m\frac{d^2 y}{dt^2} + a \frac{dy}{dt} + k(y+y_0) = F(t) + mg$$

Therefore, when the system is at equilibrium with gravity, we can ignore the gravity and

$$m\frac{d^2 y}{dt^2} + a \frac{dy}{dt} + ky = F(t)$$

Example 2..12   Suppose that the massless spring is hanging from the ceiling. When you attached a small object weighing $30g$ to its lower end, it stretched $2cm$. From this position you have stretched the spring $5cm$ and release. Find the motion of the spring.

SOLUTION By Hooke's law, we have the spring constant $k = 30/2 = 15$. Then by the Newton's 2nd law, we have

$$30\frac{d^{2}y}{dt^2} + 15y = 0$$

This is the 2nd-order linear differential equation. The characteristic equation is

$$m^2 + \frac{1}{2} = 0.$$

Solve this to get $m = \pm \frac{i}{\sqrt{2}}$. Thus,

$$y = c_{1}\cos{\frac{t}{\sqrt{2}}} + c_{2}\sin{\frac{t}{\sqrt{2}}}.$$

Now using the initial conditions $y(0) = 5, \ y^{\prime}(0) = 0$, we have

$$y = 5\cos{\frac{t}{\sqrt{2}}}\ensuremath{\ \blacksquare}$$