Lecture 15 & 16: Section 3.3 Homogeneous Equations with Constant Coefficients

General Principles

Recall that if $ y_1, y_2, \dots, y_n $ are $ n $ linearly independent solutions of an $ n^{th} $-order homogeneous equation, then the general solution is defined by the linear combination:

y = C_1 y_1 + C_2 y_2 + \dots + C_n y_n

To solve such equations, we must find $ n $ linearly independent solutions. Consider the $ n^{th} $-order homogeneous linear differential equation with constant coefficients:

a_n y^{(n)} + a_{n-1} y^{(n-1)} + \dots + a_2 y'' + a_1 y' + a_0 y = 0 \quad (1)

where the coefficients $ a_i $ are real constants and $ a_n \neq 0 $. If we let $ y = e^{rx} $, then the $ k^{th} $ derivative is $ y^{(k)} = r^k e^{rx} $. Substituting this into equation (1) leads to:

e^{rx} (a_n r^n + a_{n-1} r^{n-1} + \dots + a_1 r + a_0) = 0

Since $ e^{rx} $ is never zero, we obtain the characteristic equation:

a_n r^n + a_{n-1} r^{n-1} + \dots + a_1 r + a_0 = 0 \quad (2)

(I) Distinct Real Roots

Theorem 1: Distinct Real Roots

If the roots $ r_1, r_2, \dots, r_n $ of the characteristic equation (2) are real and distinct, then the general solution of the differential equation (1) is: $$ y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} + \dots + C_n e^{r_n x} $$

The set of functions $ \{e^{r_1 x}, e^{r_2 x}, \dots, e^{r_n x}\} $ forms a fundamental set of $ n $ linearly independent solutions.

Example 1: Initial Value Problem

Solve the following Initial Value Problem: $$ y''' + 3y'' - 10y' = 0; \quad y(0)=7, \quad y'(0)=0, \quad y''(0)=70 $$

Step 1: Characteristic equation.
$ r^3 + 3r^2 - 10r = 0 \implies r(r^2 + 3r - 10) = 0 \implies r(r+5)(r-2) = 0 $.
The roots are $ r_1 = 0, r_2 = 2, r_3 = -5 $.

Step 2: General solution.
$ y = C_1 e^{0x} + C_2 e^{2x} + C_3 e^{-5x} = C_1 + C_2 e^{2x} + C_3 e^{-5x} $.
Derivatives:
$ y' = 2C_2 e^{2x} - 5C_3 e^{-5x} $
$ y'' = 4C_2 e^{2x} + 25C_3 e^{-5x} $

Step 3: Apply initial conditions.
$ y(0) = C_1 + C_2 + C_3 = 7 $
$ y'(0) = 2C_2 - 5C_3 = 0 \implies 2C_2 = 5C_3 $
$ y''(0) = 4C_2 + 25C_3 = 70 $
Substituting $ 2C_2 = 5C_3 $ into the second derivative equation: $ 2(5C_3) + 25C_3 = 70 \implies 35C_3 = 70 \implies C_3 = 2 $.
Then $ 2C_2 = 10 \implies C_2 = 5 $.
Finally, $ C_1 + 5 + 2 = 7 \implies C_1 = 0 $.

Particular Solution: $ y(x) = 5e^{2x} + 2e^{-5x} $.

Example 2: Higher Order with Rational Root Theorem

Find a general solution of: $ y''' - 3y'' - 13y' + 15y = 0 $.

Step 1: Characteristic equation.
$ r^3 - 3r^2 - 13r + 15 = 0 $.

Algebra 2 Review: Rational Root Theorem

For a polynomial equation $ a_n r^n + a_{n-1} r^{n-1} + \dots + a_0 = 0 $ with integer coefficients, any rational root $ r = \frac{p}{q} $ in simplest form must satisfy:

$ p $ is an integer factor of the constant term $ a_0 $.
$ q $ is an integer factor of the leading coefficient $ a_n $.

Possible rational roots are factors of 15: $ \pm 1, \pm 3, \pm 5, \pm 15 $.
Testing $ r=1 $: $ 1 - 3 - 13 + 15 = 0 $. Since it works, $ (r-1) $ is a linear factor.

Step 2: Factor the polynomial.
By polynomial division: $ (r-1)(r^2 - 2r - 15) = (r-1)(r+3)(r-5) = 0 $.
The roots are $ r = 1, 5, -3 $.

General solution: $ y = C_1 e^x + C_2 e^{5x} + C_3 e^{-3x} $.

(II) Repeated Real Roots

If the characteristic equation has a repeated root $ r $ of multiplicity $ k $, we can use the method of reduction of order to show that the $ k $ linearly independent solutions corresponding to this root are:

e^{rx}, \quad xe^{rx}, \quad x^2e^{rx}, \dots, \quad x^{k-1}e^{rx}

Theorem 2: Repeated Real Roots

If the characteristic equation (2) has a repeated root $ r $ of multiplicity $ k $, then the part of the general solution of the differential equation (1) corresponding to $ r $ is of the form: $$ (C_1 + C_2 x + C_3 x^2 + \dots + C_k x^{k-1}) e^{rx} $$

Example 3: Repeated Roots

Find a general solution of: $ y^{(4)} - 8y'' + 16y = 0 $.

Characteristic equation: $ r^4 - 8r^2 + 16 = 0 \implies (r^2 - 4)^2 = 0 $.
This further factors into $ ((r+2)(r-2))^2 = (r+2)^2(r-2)^2 = 0 $.
The roots are $ r = -2 $ with multiplicity 2 and $ r = 2 $ with multiplicity 2.

General solution: $ y = (C_1 + C_2 x)e^{-2x} + (C_3 + C_4 x)e^{2x} $.

Differential Operator Notation

Define the differential operator $ D = \frac{d}{dx} $. We can represent the linear equation as $ Ly = 0 $ where: $$ L = a_n D^n + a_{n-1} D^{n-1} + \dots + a_1 D + a_0 $$ To obtain the characteristic equation, we simply replace the operator $ D $ with the variable $ r $.

Example 4: Operator Method

Find a general solution for: $ (D-1)^2(D+2)^3(D-3)y = 0 $.

Characteristic equation: $ (r-1)^2(r+2)^3(r-3) = 0 $.

$ r = 1 $ (multiplicity 2): yields solutions $ e^x, xe^x $.
$ r = -2 $ (multiplicity 3): yields solutions $ e^{-2x}, xe^{-2x}, x^2e^{-2x} $.
$ r = 3 $ (multiplicity 1): yields solution $ e^{3x} $.

General solution:
$ y = (C_1 + C_2 x)e^x + (C_3 + C_4 x + C_5 x^2)e^{-2x} + C_6 e^{3x} $.

(III) Distinct Complex Roots

Using Euler's Formula: $ e^{i\theta} = \cos\theta + i\sin\theta $. For a complex root $ r = a + bi $, the complex conjugate $ a - bi $ must also be a root of the characteristic equation.

Derivation of Real Solutions

From a conjugate pair $ r_{1,2} = a \pm bi $, the general complex solution is:
$ C_1 e^{(a+bi)x} + C_2 e^{(a-bi)x} = e^{ax} [ C_1(\cos(bx) + i\sin(bx)) + C_2(\cos(bx) - i\sin(bx)) ] $
Rearranging based on cosine and sine terms:
$ = (C_1 + C_2) e^{ax} \cos(bx) + i(C_1 - C_2) e^{ax} \sin(bx) $

We want to find two linearly independent real solutions from this complex form:

To obtain the first real solution, choose $ C_1 = C_2 = \frac{1}{2} $. This results in the real part: $ e^{ax} \cos(bx) $.
To obtain the second real solution, choose $ C_1 = \frac{-i}{2} $ and $ C_2 = \frac{i}{2} $. This results in the real part: $ e^{ax} \sin(bx) $.

Theorem 3: Distinct Complex Roots

If the characteristic equation (2) has an unrepeated pair of complex roots $ a \pm bi $ (where $ b \neq 0 $), then the corresponding part of the general solution is: $$ e^{ax}(C_1 \cos(bx) + C_2 \sin(bx)) $$

Example 6: Initial Value Problem with Complex Roots

Solve the Initial Value Problem: $ y'' - 4y' + 5y = 0 $, with $ y(0) = 1, y'(0) = 5 $.

Step 1: Characteristic equation.
$ r^2 - 4r + 5 = 0 \implies r = \frac{4 \pm \sqrt{16-20}}{2} = 2 \pm i $.
Here the real part $ a = 2 $ and imaginary coefficient $ b = 1 $.

Step 2: General solution.
$ y = e^{2x}(C_1 \cos x + C_2 \sin x) $.
Applying $ y(0) = 1 \implies C_1 = 1 $.
Finding the derivative: $ y' = 2e^{2x}(C_1 \cos x + C_2 \sin x) + e^{2x}(-C_1 \sin x + C_2 \cos x) $.
Applying $ y'(0) = 2C_1 + C_2 = 5 \implies 2(1) + C_2 = 5 \implies C_2 = 3 $.

Particular solution: $ y(x) = e^{2x}(\cos x + 3 \sin x) $.

Example 7: Complex Fourth-Order Equation

Find a general solution of $ y^{(4)} + 4y = 0 $.

Step 1: Characteristic equation.
$ r^4 + 4 = 0 \implies r^4 = -4 $.
Using polar forms: $ -4 = 4 e^{i\pi} $. Taking roots: $ r = (4 e^{i(\pi + 2n\pi)})^{1/4} $.
The four roots are $ r = \pm 1 \pm i $. These form two conjugate pairs: $ 1 \pm i $ and $ -1 \pm i $.

General solution:
$ y = e^x(C_1 \cos x + C_2 \sin x) + e^{-x}(C_3 \cos x + C_4 \sin x) $.

(IV) Repeated Complex Roots

If a conjugate pair $ a \pm bi $ has multiplicity $ k $, there are $ 2k $ linearly independent real solutions of the form:

x^p e^{ax} \cos(bx), \quad x^p e^{ax} \sin(bx) \quad \text{for } 0 \le p \le k-1

The corresponding part of the general solution is the sum:

\sum_{p=0}^{k-1} x^p e^{ax}(C_p \cos(bx) + D_p \sin(bx))

Example 8

Find a general solution of: $ (D^2 + 6D + 13)^2 y = 0 $.

Characteristic equation: $ (r^2 + 6r + 13)^2 = 0 $.
Roots of the quadratic part: $ r = \frac{-6 \pm \sqrt{36-52}}{2} = -3 \pm 2i $.
The conjugate pair $ -3 \pm 2i $ has multiplicity 2.

General solution:
$ y(x) = e^{-3x}(C_1 \cos(2x) + C_2 \sin(2x)) + xe^{-3x}(C_3 \cos(2x) + C_4 \sin(2x)) $.

Example 9: Working Backwards

Find a linear homogeneous constant-coefficient equation with the general solution: $ y(x) = Ae^{2x} + B\cos(2x) + C\sin(2x) $.

The term $ Ae^{2x} $ corresponds to the root $ r = 2 $.
The terms $ B\cos(2x) $ and $ C\sin(2x) $ correspond to the conjugate pair $ r = \pm 2i $.

Characteristic equation: $ (r-2)(r^2 + 4) = r^3 - 2r^2 + 4r - 8 = 0 $.
Differential equation: $ y''' - 2y'' + 4y' - 8y = 0 $.