Problem 11 - A First Course in Differential Equations with Modeling Applications

Question

Find the general solution of the given second-order differential equation. $$y^{\prime \prime}-4 y^{\prime}+5 y=0$$

Step-by-Step Solution

Answer

The general solution is \( y(t) = e^{2t}(C_1 \cos t + C_2 \sin t) \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( y'' - 4y' + 5y = 0 \). This is a second-order linear homogeneous differential equation with constant coefficients.

Step 2: Write the Characteristic Equation

To solve the differential equation, start by forming the characteristic equation by substituting \( y = e^{rt} \). This substitution leads to the equation \( r^2 - 4r + 5 = 0 \).

Step 3: Solve the Characteristic Equation

Solve the quadratic characteristic equation \( r^2 - 4r + 5 = 0 \) using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In this case, \( a = 1 \), \( b = -4 \), and \( c = 5 \), giving: \[ r = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{-4}}{2} = \frac{4 \pm 2i}{2} = 2 \pm i \].

Step 4: Write the General Solution

Since the characteristic roots are complex conjugates, \( 2 \pm i \), the general solution of the differential equation can be expressed as:\[ y(t) = e^{2t}(C_1 \cos t + C_2 \sin t) \]where \( C_1 \) and \( C_2 \) are arbitrary constants.

Key Concepts

Characteristic Equation

The characteristic equation is a central concept in solving linear differential equations, particularly those with constant coefficients. In the context of second-order linear homogeneous differential equations, it serves as the bridge between the differential equation itself and the algebraic methods used to solve it.

When faced with an equation like \( y'' - 4y' + 5y = 0 \), you can utilize the characteristic equation by assuming solutions of the form \( y = e^{rt} \). This assumption is designed to convert the differential equation into an algebraic one. After substitution, the differential equation simplifies to \( r^2 - 4r + 5 = 0 \), a quadratic equation, where \( r \) represents the possible rates of exponential growth or decay.

The roots of this quadratic equation, known as characteristic roots, will guide you in finding the general solution. Therefore, understanding how to derive and solve the characteristic equation is essential. It transforms the complex process of solving differential equations into the more familiar task of solving a polynomial equation.

Complex Roots

Complex roots arise when solving characteristic equations, especially when the discriminant \( b^2 - 4ac \) is negative. This results in roots of the form \( r = alpha \pm i \beta \), where \( \alpha \) and \( \beta \) are real numbers, and \( i \) is the imaginary unit.

In our example, solving \( r^2 - 4r + 5 = 0 \) with the quadratic formula leads to the characteristic roots \( 2 \pm i \). These are complex because the discriminant \( 16 - 20 \) results in \( -4 \), which contains the imaginary unit when square-rooted.

Understanding the presence of complex roots is crucial as it affects the form of the general solution. Instead of simple exponentials, the solution incorporates sine and cosine functions. This is because complex exponentials, like \( e^{(alpha + i\beta)t} \), can be expressed using Euler's formula: \( e^{i\beta t} = \cos(\beta t) + i \sin(\beta t) \). This realization allows for a general solution that involves real-valued functions even when the roots are complex.

Second-Order Linear Homogeneous Differential Equation

Second-order linear homogeneous differential equations form a substantial category of differential equations, predominantly characterized by terms up to the second derivative and homogeneity. The equation \( y'' - 4y' + 5y = 0 \) is an example, where each term contains the dependent variable or its derivatives multiplied by a constant.

These equations do not involve any multiplication of functions of \( t \), like \( f(t)y \). Hence the term "homogeneous." The structure is crucial as it determines the form of solutions—specific equations of this type can often be solved using standardized methods, notably by solving the characteristic equation.

The solutions to such equations often reflect the behavior of physical systems, such as oscillations and circuits, which is why learning to handle these equations has practical applications. Once the characteristic equation is solved, the general solution can be constructed to describe all possible solutions of the differential equation.

General Solution

The general solution of a differential equation represents the entire family of possible solutions, characterizing the behavior of the system described by the differential equation under all initial conditions.

For equations with complex roots such as \( r = 2 \pm i \), the solution is framed using both exponential and trigonometric functions. The final outcome is the expression \( y(t) = e^{2t}(C_1 \cos t + C_2 \sin t) \), where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions. This reflects the superposition principle, where the solution is a linear combination of two linearly independent functions, \( \cos t \) and \( \sin t \).

Incorporating these trigonometric functions is essential when dealing with complex roots. They ensure the solution remains real-valued despite the imaginary components. The constants \( C_1 \) and \( C_2 \) allow the solution to adapt to specific initial conditions, demonstrating the full range of dynamics the differential equation can describe.