Lecture 13: Section 3.1 Intro to 2nd-order Linear Equations

Introduction to 2nd-order Linear Equations

Recall the general forms of differential equations:

1st-order differential equation: $ G(x, y, y') = 0 $
2nd-order differential equation: $ G(x, y, y', y'') = 0 $
1st-order linear equation: $ y' + P(x)y = Q(x) $

Definition of 2nd-order Linear Differential Equation:

$$ A(x)y'' + B(x)y' + C(x)y = F(x) $$

Dividing by $ A(x) $, we obtain the standard form:

y'' + p(x)y' + q(x)y = f(x) \quad (1)

Theorem 2: Existence and Uniqueness for Linear Equations

Suppose that the functions $ p $, $ q $, and $ f $ are continuous on an open interval $ I $ containing the point $ a $. Then the Initial Value Problem:

$$ \begin{cases} y'' + p(x)y' + q(x)y = f(x) \\ y(a) = b_0, \quad y'(a) = b_1 \end{cases} $$

has a unique solution on the interval $ I $.

Example 1

Find the largest interval for which the Initial Value Problem:

$$ \begin{cases} (x-5)y'' + \ln(10-x)y' + \frac{1}{x-1}y = e^x \\ y(8) = 0, \quad y'(8) = 1 \end{cases} $$

has a unique solution.

Solution: Rewrite in standard form:

$$ y'' + \frac{\ln(10-x)}{x-5}y' + \frac{1}{(x-1)(x-5)}y = \frac{e^x}{x-5} $$

Here, $ p(x) = \frac{\ln(10-x)}{x-5} $, $ q(x) = \frac{1}{(x-1)(x-5)} $, and $ f(x) = \frac{e^x}{x-5} $.

$ p(x) $ is continuous for $ x \ne 5 $ and $ 10-x > 0 $ (which means $ x < 10 $).
$ q(x) $ is continuous for $ x \ne 1 $ and $ x \ne 5 $.
$ f(x) $ is continuous for $ x \ne 5 $.

The initial point is $ a = 8 $. The largest interval containing $ 8 $ where all functions are continuous is $ 5 < x < 10 $.

Result: (5, 10)

(I) Homogeneous 2nd-order Linear Equations

The associated homogeneous equation of standard form (1) is:

y'' + p(x)y' + q(x)y = 0 \quad (1h)

Note on "Homogeneous": In the context of higher-order linear equations, "homogeneous" means the right-hand side is zero. This is different from the "homogeneous functions" in Chapter 1 where $ \frac{dy}{dx} = F(\frac{y}{x}) $.

Chapter 1 definition: $ F(\frac{ky}{kx}) = F(\frac{y}{x}) $.
Chapter 3 definition: Equation (1h) is homogeneous because if $ y = \phi(x) $ is a solution, then $ y = c\phi(x) $ is also a solution.

Theorem 1: Principle of Superposition for Homogeneous Linear Equations

If $ y_1 $ and $ y_2 $ are two solutions of the homogeneous linear equation (1h) on an interval $ I $, then the linear combination:

$$ y = c_1y_1 + c_2y_2 $$

is also a solution on $ I $, where $ c_1 $ and $ c_2 $ are any constants.

Example 2

Verify that $ y_1 = \cos(5x) $ and $ y_2 = \sin(5x) $ are solutions of the given differential equation $ y'' + 25y = 0 $. Then find a particular solution of the form $ y = c_1y_1 + c_2y_2 $ that satisfies the initial conditions $ y(0) = 10, y'(0) = -10 $.

Verification:

$ (\cos(5x))'' + 25\cos(5x) = -25\cos(5x) + 25\cos(5x) = 0 $

$ (\sin(5x))'' + 25\sin(5x) = -25\sin(5x) + 25\sin(5x) = 0 $

General solution: $ y = c_1\cos(5x) + c_2\sin(5x) $

Derivitive: $ y' = -5c_1\sin(5x) + 5c_2\cos(5x) $

Apply $ y(0) = 10 \implies c_1\cos(0) + c_2\sin(0) = 10 \implies c_1 = 10 $

Apply $ y'(0) = -10 \implies -5(10)\sin(0) + 5c_2\cos(0) = -10 \implies 5c_2 = -10 \implies c_2 = -2 $

Particular Solution: $ y = 10\cos(5x) - 2\sin(5x) $

(II) Linear Independence and General Solutions

Definition: Two functions defined on an open interval $ I $ are said to be linearly independent on $ I $ if neither is a constant multiple of the other.

Example 3: Checking Linear Independence

(1) $ y_1 = \cos(5x), y_2 = \sin(5x) $: Linearly independent.

(2) $ y_1 = e^x, y_2 = e^{2x} $: Linearly independent.

(3) $ y_1 = \sin(2x), y_2 = \sin(x)\cos(x) $: Linearly dependent since $ y_1 = 2y_2 $.

Definition of Wronskian: For differentiable functions $ f $ and $ g $, the Wronskian is defined as:

W(f, g) = \begin{vmatrix} f & g \\ f' & g' \end{vmatrix} = fg' - f'g

Theorem 3: Wronskians of Solutions

Suppose that $ y_1 $ and $ y_2 $ are two solutions of the homogeneous 2nd-order linear equation $ y'' + p(x)y' + q(x)y = 0 $ on $ I $ where $ p $ and $ q $ are continuous.

(a) If $ y_1 $ and $ y_2 $ are linearly dependent, then $ W(y_1, y_2) = 0 $ on $ I $.

(b) If $ y_1 $ and $ y_2 $ are linearly independent, then $ W(y_1, y_2) \ne 0 $ at each point on $ I $.

Theorem 4: General Solutions of Homogeneous Equations

Let $ y_1 $ and $ y_2 $ be two linearly independent solutions of equation (1h). Then $ y = c_1y_1 + c_2y_2 $ is a general solution. The set $ \{y_1, y_2\} $ is called a fundamental set of solutions or a basis of the solution set.

(III) Linear 2nd-order Equations with Constant Coefficients

Consider the equation:

ay'' + by' + cy = 0 \quad (2h)

where $ a, b, c $ are constants. We assume a solution of the form $ y = e^{rx} $.

Then $ y' = re^{rx} $ and $ y'' = r^2e^{rx} $. Substituting into (2h):

$ ar^2e^{rx} + bre^{rx} + ce^{rx} = 0 \implies e^{rx}(ar^2 + br + c) = 0 $

Since $ e^{rx} \ne 0 $, we obtain the Characteristic Equation:

$$ ar^2 + br + c = 0 $$

Case 1: Distinct Real Roots ($ r_1 \ne r_2 $)

Theorem 5

The general solution of (2h) is $ y = c_1e^{r_1x} + c_2e^{r_2x} $.

Case 2: Repeated Real Roots ($ r_1 = r_2 = r $)

Theorem 6

The general solution of (2h) is $ y = (c_1 + c_2x)e^{rx} $. In this case, $ e^{rx} $ and $ xe^{rx} $ are two linearly independent solutions.

Example 4

Solve the Initial Value Problem: $ 9y'' - 12y' + 4y = 0 $, with $ y(0) = 3, y'(0) = 2 $.

Characteristic Equation: $ 9r^2 - 12r + 4 = (3r - 2)^2 = 0 \implies r = \frac{2}{3} $ (multiplicity 2).

General solution: $ y = (c_1 + c_2x)e^{\frac{2}{3}x} $

Derivative: $ y' = c_2e^{\frac{2}{3}x} + \frac{2}{3}(c_1 + c_2x)e^{\frac{2}{3}x} $

Apply $ y(0) = 3 \implies c_1 = 3 $

Apply $ y'(0) = 2 \implies c_2 + \frac{2}{3}(3) = 2 \implies c_2 + 2 = 2 \implies c_2 = 0 $

Particular Solution: $ y = 3e^{\frac{2}{3}x} $

Example 5

Find a homogeneous 2nd-order differential equation whose general solution is $ y = c_1 + c_2e^{-2x} $.

Solution: The roots are $ r = 0 $ and $ r = -2 $.

Characteristic equation: $ r(r + 2) = 0 \implies r^2 + 2r = 0 $

Differential Equation: $ y'' + 2y' = 0 $

Homework #51 Hint: 2nd-order Euler Equation

A 2nd-order Euler Equation is of the form:

$$ ax^2y'' + bxy' + cy = 0 $$

where $ a, b, c $ are constants. Show that if $ x > 0 $, then the substitution $ v = \ln(x) $ transforms it to a constant coefficient equation.

Hint: Let $ v = \ln(x) \implies \frac{dv}{dx} = \frac{1}{x} $.

By the chain rule: $ y' = \frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx} = \frac{1}{x} \frac{dy}{dv} $.

Second derivative: $ y'' = \frac{d}{dx}(\frac{1}{x} \frac{dy}{dv}) = -\frac{1}{x^2} \frac{dy}{dv} + \frac{1}{x} \frac{d^2y}{dv^2} \cdot \frac{dv}{dx} = -\frac{1}{x^2} \frac{dy}{dv} + \frac{1}{x^2} \frac{d^2y}{dv^2} $.

Substituting these into the Euler equation leads to a constant coefficient equation for $ y(v) $:

a \frac{d^2y}{dv^2} + (b-a) \frac{dy}{dv} + cy = 0

If $ r_1, r_2 $ are roots of this equation, the solution is $ y = c_1e^{r_1v} + c_2e^{r_2v} $.

Substituting back $ v = \ln(x) $:

y = c_1e^{r_1\ln(x)} + c_2e^{r_2\ln(x)} = c_1x^{r_1} + c_2x^{r_2}

Lecture 13 Section 3.1