ODE-05

Constant Coefficients

Constant Coefficient Linear Equations

Master the powerful techniques for solving linear ODEs with constant coefficients: characteristic equations, the superposition principle, undetermined coefficients, variation of parameters, and elegant operator methods.

Learning Objectives

Solve homogeneous equations via characteristic roots
Handle distinct, repeated, and complex roots
Apply the superposition principle
Use undetermined coefficients for particular solutions
Apply variation of parameters systematically
Master the operator method and exponential shift
Understand the Wronskian and linear independence
Solve resonance problems

1. Homogeneous Constant Coefficient Equations

Consider the $n$ -th order linear homogeneous equation with constant coefficients:

Lx := a_n x^{(n)} + a_{n-1}x^{(n-1)} + \cdots + a_1 x' + a_0 x = 0

Definition 5.1: Characteristic Equation

The characteristic equation of the homogeneous ODE is:

p(\lambda) = a_n \lambda^n + a_{n-1}\lambda^{n-1} + \cdots + a_1\lambda + a_0 = 0

The roots $\lambda_1, \ldots, \lambda_m$ (with multiplicities $n_1, \ldots, n_m$ ) are called the characteristic values or eigenvalues of the equation.

Theorem 5.1: General Solution of Homogeneous Constant Coefficient ODE

Let the characteristic equation have $m$ distinct roots $\lambda_1, \ldots, \lambda_m$ with multiplicities $n_1, \ldots, n_m$ where $\sum n_k = n$ . Then:

$\{t^j e^{\lambda_k t} : 1 \leq k \leq m, 0 \leq j < n_k\}$ forms a set of $n$ linearly independent solutions
The general solution is:
$x(t) = \sum_{k=1}^{m} \sum_{j=0}^{n_k-1} C_{jk} t^j e^{\lambda_k t}$

Proof of Theorem 5.1:

Step 1: Show that $t^j e^{\lambda_k t}$ are solutions. Since $p(\lambda_k) = 0$ with multiplicity $n_k$ , we have $p(\lambda_k) = p'(\lambda_k) = \cdots = p^{(n_k-1)}(\lambda_k) = 0$ .

Using the identity $L(t^j e^{\lambda t}) = \frac{\partial^j}{\partial \lambda^j}L(e^{\lambda t})|_{\lambda=\lambda_k} = \frac{\partial^j}{\partial \lambda^j}(p(\lambda)e^{\lambda t})|_{\lambda=\lambda_k}$ :

L(t^j e^{\lambda_k t}) = \sum_{l=0}^{j} \binom{j}{l} p^{(l)}(\lambda_k) t^{j-l} e^{\lambda_k t} = 0 \text{ for } j < n_k

Step 2: Prove linear independence. Suppose $\sum_{k,j} C_{jk} t^j e^{\lambda_k t} \equiv 0$ . We show $C_,{jk}, = 0$ for all $j, k$ .

Multiply by $e^{-\lambda_m t}$ and differentiate repeatedly to eliminate terms, reducing to the case of fewer distinct roots. By induction, all coefficients vanish. ∎

Summary: Root Types and Solutions

Distinct Real Roots $\lambda_1, \lambda_2, \ldots$ :

Solutions: $e^{\lambda_1 t}, e^{\lambda_2 t}, \ldots$

Repeated Real Root $\lambda$ of multiplicity $k$ :

Solutions: $e^{\lambda t}, te^{\lambda t}, t^2 e^{\lambda t}, \ldots, t^{k-1}e^{\lambda t}$

Complex Conjugate Roots $\alpha \pm i\beta$ :

Real Solutions: $e^{\alpha t}\cos\beta t, e^{\alpha t}\sin\beta t$

Example 5.1: Solving a Homogeneous Equation

Problem: Find the general solution of $x^,{(5)}, - 3x^,{(4)}, + 4x''' - 4x'' + 3x' - x = 0$ .

Solution:

The characteristic equation is:

\lambda^5 - 3\lambda^4 + 4\lambda^3 - 4\lambda^2 + 3\lambda - 1 = 0

Factoring: $(\lambda - 1)^3(\lambda^2 + 1) = 0$

Roots: $\lambda_1 = 1$ (multiplicity 3), $\lambda_2 = i, \lambda_3 = -i$ (simple)

From $\lambda_1 = 1$ (triple): $e^t, te^t, t^2 e^t$

From $\lambda = \pm i$ : $\cos t, \sin t$

General solution:

x(t) = (C_1 + C_2 t + C_3 t^2)e^t + C_4 \cos t + C_5 \sin t

2. Superposition Principle

Theorem 5.2: Superposition Principle

For the linear operator $L$ and nonhomogeneous equation $Lx = f(t)$ :

\text{General Solution} = \text{Homogeneous Solution} + \text{Particular Solution}

If $x_h$ is the general solution of $Lx = 0$ and $x_p$ is any particular solution of $Lx = f(t)$ , then $x = x_h + x_p$ is the general solution of the nonhomogeneous equation.

Definition 5.2: Wronskian

For $n$ functions $\phi_1, \ldots, \phi_n$ , the Wronskian is:

W(t) = W(\phi_1, \ldots, \phi_n)(t) = \begin{vmatrix} \phi_1 & \phi_2 & \cdots & \phi_n \\ \phi_1' & \phi_2' & \cdots & \phi_n' \\ \vdots & \vdots & \ddots & \vdots \\ \phi_1^{(n-1)} & \phi_2^{(n-1)} & \cdots & \phi_n^{(n-1)} \end{vmatrix}

If $W(t) \neq 0$ for some $t$ , the functions are linearly independent.

3. Method of Undetermined Coefficients

For specific forms of $f(t)$ , we can guess the form of a particular solution and determine its coefficients.

Algorithm 5.1: Undetermined Coefficients Method

Input: $Lx = f(t)$ where $f(t) = P_l(t)e^,{\mu t}$ (polynomial × exponential)

Output: Particular solution $x_p$

1. Find the characteristic roots of $Lx = 0$

2. If $\mu$ is not a characteristic root:

Try $x_p = Q_l(t)e^,{\mu t}$ where $Q_l$ is a polynomial of degree $l$

3. If $\mu$ is a characteristic root of multiplicity $k$ :

Try $x_p = t^k Q_l(t)e^,{\mu t}$

4. Substitute into the equation and solve for coefficients

5. return $x_p$

Remark:

For trigonometric forcing $f(t) = (P_l \cos\beta t + Q_m \sin\beta t)e^{\alpha t}$ , try $x_p = (C_s \cos\beta t + D_s \sin\beta t)e^{\alpha t}$ where $s = \max(l, m)$ . If $\alpha + i\beta$ is a characteristic root of multiplicity $k$ , multiply by $t^k$ .

Example 5.2: Undetermined Coefficients

Problem: Solve $x''' + 3x'' + 3x' + x = e^,{-t},(t + 5) + e^t + \sin t + 1$ .

Solution:

Characteristic equation: $(\lambda + 1)^3 = 0$ , so $\lambda = -1$ (triple root).

Homogeneous solution: $x_h = (C_1 + C_2 t + C_3 t^2)e^,{-t}$

For $f_1 = e^,{-t},(t + 5)$ : Since $-1$ is a triple root, try $x_1 = t^3(at + b)e^,{-t}$ .

Substituting: $24bt^3 + 24a = t + 5$ gives $a = 5/24, b = 1/24$ .

x_1 = \frac{t^3}{24}(t + 20)e^{-t}

For $f_2 = e^t$ : Try $x_2 = Ae^t$ . Substituting: $8A = 1$ , so $A = 1/8$ .

For $f_3 = \sin t$ : Try $x_3 = a\cos t + b\sin t$ . Solving: $a = b = -1/4$ .

For $f_4 = 1$ : Try $x_4 = A$ . Clearly $A = 1$ .

General solution:

x = (C_1 + C_2 t + C_3 t^2)e^{-t} + \frac{t^3(t+20)}{24}e^{-t} + \frac{1}{8}e^t - \frac{1}{4}(\sin t + \cos t) + 1

4. Variation of Parameters

Theorem 5.3: Variation of Parameters for Second Order

If $\phi_1, \phi_2$ are linearly independent solutions of $x'' + p(t)x' + q(t)x = 0$ , then a particular solution of $x'' + p(t)x' + q(t)x = f(t)$ is:

x_p(t) = \int_{t_0}^{t} \frac{\phi_1(s)\phi_2(t) - \phi_1(t)\phi_2(s)}{W(s)} f(s)\,ds

where $W(s) = \phi_1(s)\phi_2'(s) - \phi_1'(s)\phi_2(s)$ is the Wronskian.

Example 5.3: Variation of Parameters

Problem: Solve $x'' + x = \sec t$ .

Solution:

Homogeneous solutions: $\phi_1 = \cos t, \phi_2 = \sin t$

Wronskian: $W = \cos t \cdot \cos t - (-\sin t) \cdot \sin t = 1$

Using the formula:

x_p = \int (\cos s \sin t - \cos t \sin s) \sec s\,ds

= \sin t \int \cos s \sec s\,ds - \cos t \int \sin s \sec s\,ds

= \sin t \cdot t - \cos t \cdot (-\ln|\cos t|)

= t\sin t + \cos t \ln|\cos t|

5. Operator Method

Definition 5.3: Differential Operator

Define the differential operator $D = d/dt$ and the operator polynomial:

P(D) = a_n D^n + a_{n-1}D^{n-1} + \cdots + a_1 D + a_0

The ODE becomes $P(D)x = f(t)$ , and formally, $x = \frac,{1}{P(D)}, f(t)$ .

Theorem 5.4: Operator Properties

1. $P(D)e^,{at}, = P(a)e^,{at}$

2. Exponential Shift: $P(D)(e^,{at},u(t)) = e^,{at},P(D+a)u(t)$

3. $P(D^2)\sin\omega t = P(-\omega^2)\sin\omega t$ and similarly for $\cos\omega t$

Algorithm 5.2: Operator Method for Particular Solutions

Case 1: $f(t) = e^,{at}$

If $P(a) \neq 0$ : $x_p = \frac,{1}{P(a)},e^,{at}$

If $a$ is a root of multiplicity $k$ : $x_p = \frac,{t^k}{P^{(k)}(a)},e^,{at}$

Case 2: $f(t) = P_n(t)$ (polynomial)

Expand $\frac,{1}{P(D)}$ as power series in $D$

Case 3: $f(t) = e^,{at},g(t)$

Use $\frac,{1}{P(D)},e^,{at},g(t) = e^,{at},\frac,{1}{P(D+a)},g(t)$

Example 5.4: Operator Method

Problem: Find a particular solution of $x'' - 2x' + x = te^t$ .

Solution:

The equation is $(D-1)^2 x = te^t$ . Using the exponential shift:

x_p = \frac{1}{(D-1)^2}(te^t) = e^t \frac{1}{D^2}(t) = e^t \int\int t\,dt\,dt

= e^t \cdot \frac{t^3}{6}

Practice Quiz

Constant Coefficient Equations Quiz

Questions

Correct

Accuracy

The characteristic equation of

x'' - 5x' + 6x = 0

is:

Easy

Not attempted

If the characteristic equation has roots

\lambda_1 = 2

and

\lambda_2 = 3

, the general solution is:

Easy

Not attempted

\lambda = 2

is a double root of the characteristic equation, the general solution is:

Easy

Not attempted

For complex roots

\lambda = \alpha \pm i\beta

, the real solutions are:

Medium

Not attempted

The Wronskian of

e^{2t}

and

e^{3t}

is:

Medium

Not attempted

For

x'' + x = e^t

, a particular solution using undetermined coefficients has the form:

Medium

Not attempted

For

x'' - 4x = e^{2t}

, the particular solution form is:

Medium

Not attempted

In the operator method,

D

represents:

Medium

Not attempted

The operator identity

P(D)(e^{at}u(t)) =

Hard

Not attempted

For finding a particular solution of

x'' + 4x = \sin 2t

(resonance case), the correct form is:

Hard

Not attempted

Frequently Asked Questions

How do I handle complex characteristic roots?

For complex conjugate roots

\alpha \pm i\beta

, the complex solutions

e^,{(\alpha \pm i\beta)t}

give real solutions

e^,{\alpha t},\cos\beta t

and

e^,{\alpha t},\sin\beta t

. For repeated complex roots, multiply by powers of

t

as with real roots.

When should I use undetermined coefficients vs. variation of parameters?

Use undetermined coefficients when

f(t)

is a polynomial, exponential, sine/cosine, or products thereof. It's faster when applicable. Use variation of parameters for any

f(t)

—it always works but may involve difficult integrals.

What is resonance?

Resonance occurs when the forcing frequency matches a natural frequency of the system. For

x'' + \omega^2 x = A\cos\omega t

, the characteristic roots are

\pm i\omega

, so

\cos\omega t

is a homogeneous solution. The particular solution has the form

t(B\cos\omega t + C\sin\omega t)

, growing linearly in time.

How does the operator method simplify calculations?

The operator method treats derivatives algebraically. For example,

P(D)e^,{at}, = P(a)e^,{at}

instantly gives particular solutions for exponential forcing. The exponential shift

P(D)(e^,{at},u) = e^,{at},P(D+a)u

reduces problems to simpler forms.

Why is the Wronskian important?

The Wronskian

W

determines linear independence of solutions: if

W \neq 0

at any point, the solutions are independent. It also appears in variation of parameters formulas and satisfies Abel's identity:

W(t) = W(t_0)e^{-\int_{t_0}^{t} p(s)ds}

for the equation

x'' + p(t)x' + q(t)x = 0

Can these methods handle variable coefficient equations?

Characteristic equation and undetermined coefficients are specific to constant coefficients. Variation of parameters works for any linear equation if you know the homogeneous solutions. For variable coefficients, other methods (power series, Frobenius, etc.) are typically needed to find homogeneous solutions first.