Problem 57 - Advanced Engineering Mathematics

Question

Find a homogeneous linear differential equation with constant coefficients whose general solution is given. $$ y \quad c_{1}+c_{2} x+c_{3} e^{7 x} $$

Step-by-Step Solution

Answer

The differential equation is \( y''' - 7y'' = 0 \).

Step 1: Identify the General Solution Components

The general solution given is \( y = c_{1} + c_{2} x + c_{3} e^{7x} \). It comprises a polynomial component \( c_{1} + c_{2} x \) and an exponential component \( c_{3} e^{7x} \). These components will guide us to determine the roots of the characteristic equation of the differential equation.

Step 2: Determine Characteristics for Polynomial Component

The polynomial component \( c_{1} + c_{2} x \) reflects the presence of real roots with multiplicity. Specifically, it corresponds to a root \( r = 0 \) with multiplicity 2 in the characteristic equation, resulting in solutions like \( c_{1} + c_{2} x \).

Step 3: Determine Characteristics for Exponential Component

The exponential term \( c_{3} e^{7x} \) reflects the existence of a real, distinct root \( r = 7 \) in the characteristic equation. This term indicates an exponential solution linked to this non-zero real root.

Step 4: Formulate the Characteristic Polynomial

Based on the roots identified: a double root at \( r = 0 \) and a simple root at \( r = 7 \). The characteristic polynomial can be expressed as \((r - 0)^2 (r - 7) = r^2 (r - 7)\).

Step 5: Write the Differential Equation

The characteristic polynomial \( r^2 (r - 7) \) corresponds to the differential operator \( (D^2)(D - 7) \), where \( D = \frac{d}{dx} \). Applying this gives the differential equation: \[ \frac{d^3y}{dx^3} - 7 \frac{d^2y}{dx^2} = 0. \]

Key Concepts

Characteristic Equation

In the study of homogeneous linear differential equations, the characteristic equation plays a critical role. Basically, it is a tool that helps to find solutions for differential equations.
The characteristic equation is obtained from the differential equation by substituting the derivatives with powers of a variable (usually denoted as \( r \)). This substitution transforms the differential operator into a polynomial equation. In our exercise, we identified the characteristic polynomial as \( r^2 (r - 7) \) from the general solution components.

• Step 2 of the solution revealed real roots indicating polynomial components.
• Step 3 connected exponential solutions with distinct roots.

Understanding the characteristic equation helps in determining the nature of solutions, whether they are exponential or polynomial, which directly ties to the roots of this equation.

Roots of Polynomial

The roots of the characteristic polynomial are crucial in forming the solution to a differential equation. They determine the form of the general solution by dictating whether the solution involves exponentials, sines, or polynomials.

In our case, the solution consists of a polynomial component \( c_1 + c_2 x \) and an exponential component \( c_3 e^{7x} \).

• The polynomial component arises from a double root at \( r = 0 \).
• The exponential component is due to a single, distinct root of \( r = 7 \).

These roots guide us in constructing the characteristic polynomial \( (r - 0)^2 (r - 7) = r^2 (r - 7) \). Knowing the multiplicity and values of these roots is essential when forming the differential equation and understanding the type of solution it produces.

General Solution

The general solution of a homogeneous linear differential equation encompasses all possible solutions derived from its characteristic equation. It typically comprises terms based on the roots and their multiplicities.

For the given solution \( y = c_1 + c_2 x + c_3 e^{7x} \), each part originates from the roots of the characteristic polynomial.

• \( c_1 + c_2 x \) corresponds to the double root \( r = 0 \).
• \( c_3 e^{7x} \) stems from the distinct root \( r = 7 \).

Thus, the general solution is a combination of these terms, accounting for all the contributions from each root. It's the blueprint for constructing specific solutions when initial or boundary conditions are applied.

Exponential Functions

Exponential functions often appear in the solutions of linear differential equations, especially when the characteristic equation has distinct or complex roots. They take the form \( e^{rx} \), where \( r \) is a root from the characteristic polynomial.

In our example, the term \( c_3 e^{7x} \) represents the exponential part of the solution.

• It arises from the distinct root \( r = 7 \).
• This exponential solution contributes its unique shape to the behavior of the general solution.

Understanding how exponential functions integrate into the general solution helps predict the behavior of the system modeled by the differential equation. Exponentials reflect growth or decay processes, depending on whether their root is positive or negative.