Problem 50 - Advanced Engineering Mathematics

Question

In Problems 49-58 find a homogeneous linear differential equation with constant coefficients whose general solution is given. $$ y \quad c_{1} e^{-5 x}+c_{2} e^{-4 x} $$

Step-by-Step Solution

Answer

The differential equation is \( y'' + 9y' + 20y = 0 \).

Step 1: Identify the General Solution Form

The given general solution is \( y = c_1 e^{-5x} + c_2 e^{-4x} \). This implies the differential equation has solutions in terms of exponential functions of the form \( e^{rx} \) where \( r = -5 \) and \( r = -4 \).

Step 2: Determine the Characteristic Equation

For each root \( r \) in the general solution, there corresponds a factor \( (r + 5) \) and \( (r + 4) \) from the characteristic equation. Thus, the characteristic equation for this differential equation is \((r + 5)(r + 4) = 0\).

Step 3: Expand the Characteristic Equation

Expanding \((r + 5)(r + 4) = 0\) gives \( r^2 + 9r + 20 = 0 \). This is the characteristic equation for our differential equation.

Step 4: Formulate the Differential Equation

The characteristic equation \( r^2 + 9r + 20 = 0 \) translates to a differential equation with constant coefficients as \( y'' + 9y' + 20y = 0 \).

Key Concepts

Characteristic Equation

In the context of differential equations, the characteristic equation is a crucial concept that helps us unlock the general solution of a given equation. When we are asked to find a differential equation with a specified solution, the characteristic equation plays an essential role. It is derived from the given general solution, which often involves exponential functions. For homogeneous linear differential equations with constant coefficients, the solution includes terms like \( e^{rx} \), where \( r \) represents roots of the characteristic equation.

• For every term \( c_1 e^{-5x} \) and \( c_2 e^{-4x} \) in the solution, there's a root \( r \) that contributes to the characteristic equation.
• In our case, \( r = -5 \) and \( r = -4 \), which provide the factors \((r + 5)\) and \((r + 4)\).

Therefore, multiplying these factors gives us the characteristic equation \((r + 5)(r + 4) = 0\). Expanding this multiplication leads to \( r^2 + 9r + 20 = 0 \). Understanding how to form and expand this equation is fundamental to solving homogeneous linear differential equations.

Constant Coefficients

Differential equations with constant coefficients have the form \( a_n y^{(n)} + a_{n-1} y^{(n-1)} + ... + a_1 y' + a_0 y = 0 \), where each coefficient \( a_i \) is constant. This particular structure allows for solutions that are exponential functions, which can be more manageable compared to variable coefficient equations.

• The example given shows coefficients as part of the differential equation \( y'' + 9y' + 20y = 0 \).
• Each term of this equation stems from the expanded characteristic equation \( r^2 + 9r + 20 = 0 \).

The steady nature of constant coefficients tends to simplify the solving process, as it standardizes the way solutions and characteristic equations form. This regularity is what makes this category of differential equations more approachable and easier to learn.

General Solution

The general solution of a differential equation represents all possible solutions. For linear differential equations with constant coefficients, these solutions are typically expressed in terms of exponential functions. When given a general solution, like \( y = c_1 e^{-5x} + c_2 e^{-4x} \), each exponential term corresponds to a particular root of the characteristic equation.

• The solution combines these exponential terms, each multiplied by an arbitrary constant \( c_1 \) or \( c_2 \).
• The constants are determined based on initial conditions or boundary values when solving specific scenarios.

Understanding general solutions is crucial for predicting the behavior of systems modeled by differential equations. The flexibility provided by the arbitrary constants \( c_1 \) and \( c_2 \) accommodates various initial conditions and unique real-world situations.

Exponential Functions

Exponential functions are at the core of solving homogeneous linear differential equations, specifically those with constant coefficients. These functions, of the form \( e^{rx} \), naturally arise from the characteristic equation. When solving these differential equations, each root \( r \) of the characteristic equation indicates a possible exponential solution.

• In our example, roots \( r = -5 \) and \( r = -4 \) give us the exponential terms \( e^{-5x} \) and \( e^{-4x} \) respectively.
• These terms combine linearly, as in \( c_1 e^{-5x} + c_2 e^{-4x} \), to form the general solution.

The use of exponential functions simplifies the solutions of differential equations because they have neat properties for differentiation. This yields solutions that are both elegant and practical for modeling and analyzing continuous systems.