Problem 7 - Calculus 3

Question

For each of the following problems, verify that the given function is a solution to the differential equation. Use a graphing utility to graph the particular solutions for several values of \(c_{1}\) and \(c_{2} .\) What do the solutions have in common? $$ [\mathrm{T}] \mathrm{y}^{\prime \prime}+2 y^{\prime}-3 y=0 ; \quad y(x)=c_{1} e^{x}+c_{2} e^{-3 x} $$

Step-by-Step Solution

Answer

The function satisfies the differential equation and solutions exhibit exponential growth or decay.

Step 1: Compute the Derivatives

First, compute the first and second derivatives of the given function. The function is: \[ y(x) = c_1 e^{x} + c_2 e^{-3x} \] The first derivative is:\[ y'(x) = \frac{d}{dx}[c_1 e^{x} + c_2 e^{-3x}] = c_1 e^{x} - 3c_2 e^{-3x} \]The second derivative is:\[ y''(x) = \frac{d}{dx}[c_1 e^{x} - 3c_2 e^{-3x}] = c_1 e^{x} + 9c_2 e^{-3x} \]

Step 2: Substitute into the Differential Equation

Substitute the computed derivatives and the function into the given differential equation \[ y'' + 2y' - 3y = 0 \]. Substituting, we have:\[ c_1 e^{x} + 9c_2 e^{-3x} + 2(c_1 e^{x} - 3c_2 e^{-3x}) - 3(c_1 e^{x} + c_2 e^{-3x}) \]

Step 3: Simplify the Expression

Simplify the expression after substitution:\[(c_1 e^{x} + 9c_2 e^{-3x}) + 2c_1 e^{x} - 6c_2 e^{-3x} - 3c_1 e^{x} - 3c_2 e^{-3x}\]Combine like terms:\[(1 + 2 - 3)c_1 e^{x} + (9 - 6 - 3)c_2 e^{-3x} = 0c_1 e^{x} + 0c_2 e^{-3x} = 0.\]Since both parts equal zero, the function satisfies the differential equation.

Step 4: Discuss the General Solutions

The family of solutions \( y(x) = c_1 e^x + c_2 e^{-3x} \) forms a general solution to the differential equation. The solutions have the exponential terms \( e^x \) and \( e^{-3x} \), which are characteristic of the roots of the characteristic equation of the differential equation.

Step 5: Analyze Graphs of Particular Solutions

Use a graphing utility to graph \( y(x) = c_1 e^x + c_2 e^{-3x} \) for several values of \( c_1 \) and \( c_2 \). Observe that all graphs will tend towards the behavior dictated by the exponents \( e^x \) and \( e^{-3x} \) as \( x \) increases or decreases, showing exponential growth or decay patterns.

Key Concepts

General Solutions

Differential equations often seek to relate a function with its derivatives, creating expressions that predict behavior over time or space. The general solution of a differential equation includes every possible solution, incorporating constants that can adjust to initial conditions or specific scenarios.
The exercise at hand involves a second-order linear homogeneous differential equation:

• It is given by: \( y'' + 2y' - 3y = 0 \).
• One general solution is: \( y(x) = c_1 e^x + c_2 e^{-3x} \).

The general solution covers all specific instances when parameters like initial values are applied. By including constants \( c_1 \) and \( c_2 \), the formula represents an entire set of potential solutions. These constants can be adapted to match specific initial or boundary conditions, tailoring the general solution to particular cases. As a result, you can generate specific solutions by assigning different values to these constants. This flexibility is foundational in solving real-world problems using differential equations.

Exponential Functions

Exponential functions play a crucial role in solving differential equations. They frequently appear in the solutions of linear differential equations with constant coefficients. In our example:

• The terms \( e^x \) and \( e^{-3x} \) are exponential functions.
• Exponential functions are characteristic because they describe natural growth or decay.

In the context of differential equations, exponential functions describe how solutions grow or shrink as the independent variable (often time or space) changes.
The significant advantage of exponential functions in these solutions is their elegant properties; they differentiate to themselves, making calculations simpler. When \( x \) increases:

• \( e^x \) leads to exponential growth.
• \( e^{-3x} \) leads to exponential decay due to the negative coefficient.

This behavior represents dynamic systems, such as population growth or radioactive decay, providing a natural framework for modeling.

Characteristic Equation

The characteristic equation is a pivotal concept for solving linear homogeneous differential equations with constant coefficients. It provides a way to determine the possible forms of the solution involving exponential functions. Here's how it works:

• For the given equation \( y'' + 2y' - 3y = 0 \), the associated characteristic equation is \( r^2 + 2r - 3 = 0 \).

To solve the characteristic equation, we find the roots, which determine the exponents in the solution:

• Using the quadratic formula or factoring, you determine the roots of the equation.
• For this equation, the roots are \( r = 1 \) and \( r = -3 \).

These roots indicate that the general solution comprises exponential functions: \( e^{1x} = e^x \) and \( e^{-3x} \). Multiple types of roots (real, repeated, complex) affect the form of the solution. Hence, the characteristic equation simplifies finding the exact exponential terms, guiding towards the general solution of the differential equation.