Problem 13 - Advanced Engineering Mathematics

Question

In Problems 13-28, use the procedures developed in this chapter to find the general solution of each differential equation. $$ y^{\prime \prime}-2 y^{\prime}-2 y=0 $$

Step-by-Step Solution

Answer

The general solution is \( y(t) = C_1 e^{(1+\sqrt{3})t} + C_2 e^{(1-\sqrt{3})t} \).

Step 1: Identify the Type of Differential Equation

The given differential equation \( y'' - 2y' - 2y = 0 \) is a second-order linear homogeneous differential equation with constant coefficients.

Step 2: Write the Characteristic Equation

For a second-order linear homogeneous differential equation of the form \( a y'' + b y' + c y = 0 \), the characteristic equation is \( ar^2 + br + c = 0 \). Here, the characteristic equation is \( r^2 - 2r - 2 = 0 \).

Step 3: Solve the Characteristic Equation

Solve the quadratic equation \( r^2 - 2r - 2 = 0 \) using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -2, c = -2 \).

Step 4: Calculate the Roots

Substitute the values into the quadratic formula: \( r = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)} = \frac{2 \pm \sqrt{4 + 8}}{2} \). Simplifying gives \( r = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3} \). The roots are \( r_1 = 1 + \sqrt{3} \) and \( r_2 = 1 - \sqrt{3} \).

Step 5: Write the General Solution

For distinct real roots \( r_1 \) and \( r_2 \), the general solution of the differential equation is \( y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \). Substitute the roots: \( y(t) = C_1 e^{(1+\sqrt{3})t} + C_2 e^{(1-\sqrt{3})t} \).

Key Concepts

Second-Order Linear Differential Equations

Differential equations are equations that involve derivatives of a function. A second-order linear differential equation involves the second derivative, denoted as \( y'' \), and is typically written in the standard form \( a y'' + b y' + c y = 0 \), where \( a \), \( b \), and \( c \) are constants.

These equations are "linear" because they are linear in terms of the function \( y \) and its derivatives. "Second-order" indicates the highest derivative in the equation is the second derivative.
This type of differential equation is important in fields like physics and engineering, where they often model systems experiencing acceleration or wave propagation.

To solve these equations, especially when they are homogeneous with constant coefficients, we use the characteristic equation method.

Characteristic Equation

When dealing with a second-order linear homogeneous differential equation, the first step is to convert it into a characteristic equation. The characteristic equation is a quadratic equation derived from the differential equation's coefficients.

Given a differential equation \( a y'' + b y' + c y = 0 \), its characteristic equation is formed as \( ar^2 + br + c = 0 \).
This transformation allows for finding the roots, \( r \), which lead directly to the solution of the original differential equation.

The characteristic equation is crucial because the nature of its roots (real, repeated, or complex) determines the form of the solution, unlocking the behavior of the system described by the differential equation.

Quadratic Formula

The quadratic formula is a fundamental tool for solving the characteristic equation, which is always quadratic in nature. This formula is given by: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation derived in the characteristic step. The \( \pm \) symbol indicates there are generally two roots. These roots can be real, repeated, or complex, influenced by the discriminant \( b^2 - 4ac \):

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is one repeated real root.
• If it is negative, the roots are complex conjugates.

This formula is essential as it facilitates finding the roots systematically, thereby assisting in solving the differential equation.

General Solution

Once the roots of the characteristic equation are found, the next step is to formulate the general solution of the differential equation. The form of this solution depends on the nature of the roots:

• For two distinct real roots \( r_1 \) and \( r_2 \), the general solution is given by:\[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \]
• For a repeated real root \( r \), it is:\[ y(t) = (C_1 + C_2 t) e^{r t} \]
• With complex roots \( \alpha \pm \beta i \), the solution takes the form:\[ y(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t)) \]

The constants \( C_1 \) and \( C_2 \) are determined by initial conditions or boundary conditions, which tailor the general solution to fit specific scenarios. The general solution provides a "family" of solutions encompassing all possible states of the system.