Problem 14 - Calculus and its applications

Question

Let \(y^{\prime \prime}-7 y^{\prime}-44 y=0\). a) Show that \(y=e^{11 x}\) is a solution of this differential equation. b) Show that \(y=e^{-4 x}\) is a solution. c) Show that \(y=C_{1} e^{11 x}+C_{2} e^{-4 x}\) is a solution, where \(C_{1}\) and \(C_{2}\) are constants.

Step-by-Step Solution

Answer

a) Yes, it is a solution. b) Yes, it is a solution. c) Yes, it is a solution.

Step 1: Confirm Solution for y=e^{11x}

First, find the derivative of \(y = e^{11x}\). The first derivative is \(y' = 11e^{11x}\), and the second derivative is \(y'' = 121e^{11x}\). Substitute these into the differential equation \(y'' - 7y' - 44y = 0\):\[121e^{11x} - 7(11e^{11x}) - 44e^{11x} = 0\]Simplifying:\[121e^{11x} - 77e^{11x} - 44e^{11x} = 0\]\[(121 - 77 - 44)e^{11x} = 0\]\[0 = 0\]Since this equation holds true, \(y = e^{11x}\) is indeed a solution.

Step 2: Confirm Solution for y=e^{-4x}

Find derivatives of \(y = e^{-4x}\). The first derivative is \(y' = -4e^{-4x}\), and the second derivative is \(y'' = 16e^{-4x}\). Substitute into the differential equation:\[16e^{-4x} - 7(-4e^{-4x}) - 44e^{-4x} = 0\]Simplifying:\[16e^{-4x} + 28e^{-4x} - 44e^{-4x} = 0\]\[(16 + 28 - 44)e^{-4x} = 0\]\[0 = 0\]This confirms that \(y = e^{-4x}\) is also a solution.

Step 3: Confirm General Solution for y=C_1e^{11x}+C_2e^{-4x}

Calculate the derivatives of \(y = C_1e^{11x} + C_2e^{-4x}\). The first derivative is \(y' = 11C_1e^{11x} - 4C_2e^{-4x}\) and the second derivative is \(y'' = 121C_1e^{11x} + 16C_2e^{-4x}\). Substitute these into the differential equation:\[(121C_1e^{11x} + 16C_2e^{-4x}) - 7(11C_1e^{11x} - 4C_2e^{-4x}) - 44(C_1e^{11x} + C_2e^{-4x}) = 0\]Distribute:\[121C_1e^{11x} + 16C_2e^{-4x} - 77C_1e^{11x} + 28C_2e^{-4x} - 44C_1e^{11x} - 44C_2e^{-4x} = 0\]Combine like terms:\[(121 - 77 - 44)C_1e^{11x} + (16 + 28 - 44)C_2e^{-4x} = 0\]This simplifies to:\[0 = 0\]Thus, \(y = C_1e^{11x} + C_2e^{-4x}\) is indeed a solution.

Key Concepts

Second Order Linear Differential Equations

A second order linear differential equation involves derivatives up to the second order and has the general form: \[a y'' + b y' + c y = 0\] where \(a\), \(b\), and \(c\) are constants.

This particular type of equation is characterized by the highest derivative, which is the second derivative (\(y''\)). Linear means that the terms in \(y\), \(y'\), and \(y''\) are only multiplied by constants and not by each other.

• The coefficients of the different derivatives (\(a\), \(b\), \(c\)) indicate the type and nature of the solution.
• The way these coefficients interact affects whether solutions are oscillatory, exponential, or a combination of both.

Second order differential equations are vital in modeling behaviors like mechanical vibrations and electrical circuits. Understanding them requires recognizing the structure and solving using systematic approaches like substitution or characteristic equations.

Solution Verification

The process of solution verification involves substituting proposed solutions into the original differential equation to ensure they satisfy it. This means we calculate derivatives of the proposed solutions and check if they fulfill \[y'' - 7y' - 44y = 0\]when substituted.

In the given exercise, we checked for two specific solutions: \( y = e^{11x} \) and \( y = e^{-4x} \). We calculated derivatives, substituted, and confirmed

• For \( y = e^{11x} \), its derivatives satisfy the equation because all terms canceled, resulting in \(0 = 0\).
• Similarly, \( y = e^{-4x} \) met the same condition, affirming it as a valid solution.

For the combination solution \(y = C_1e^{11x} + C_2e^{-4x}\), a similar substitution and simplification showed it satisfies the equation over any values of \(C_1\) and \(C_2\). Thus, ensuring all possible solutions are verified provides confidence in their accuracy.

Exponential Functions

Exponential functions take the form \(y = e^{kx}\), where \(e\) represents Euler’s number \(\approx 2.718\). These functions can either rapidly grow or decay based on the sign of \(k\).

In the context of differential equations:

• They are easy to differentiate, making them convenient for substitution and verification processes.
• The exponential nature helps describe various physical phenomena: population growth, radioactive decay, and temperature changes over time.

In this lesson, two exponential functions, \(e^{11x}\) and \(e^{-4x}\), serve as solutions for the differential equation. Their behavior—growth vs. decay—depends on the sign of \(k\): positive for growth (like \(11\)), and negative for decay (like \(-4\)). Such solutions are prevalent in modeling real-life gradual processes. With their predictable pattern of change, exponential functions play a crucial role in understanding time-related behaviors in science and engineering.