Problem 69 - A First Course in Differential Equations with Modeling Applications

Question

Use a CAS as an aid in solving the auxiliary equation. Form the general solution of the differential equation. Then use a CAS as an aid in solving the system of equations for the coefficients \(c_{i}, i=1,2,3,4\) that results when the initial conditions are applied to the general solution. $$\begin{array}{l} 2 y^{(4)}+3 y^{\prime \prime \prime}-16 y^{\prime \prime}+15 y^{\prime}-4 y=0 \\\ y(0)=-2, y^{\prime}(0)=6, y^{\prime \prime}(0)=3, y^{\prime \prime \prime}(0)=\frac{1}{2} \end{array}$$

Step-by-Step Solution

Answer

General solution: \(y(t) = e^{2t} - 3e^{-t} + 8e^{\frac{1}{2}t}\sin(\frac{\sqrt{7}}{2}t)\). Coefficients: \(c_1 = 1, c_2 = -3, c_3 = 0, c_4 = 8\).

Step 1: Define the Auxiliary Equation

For the given differential equation, the auxiliary equation is formed by replacing derivatives of \(y\) with the powers of \(r\). So the equation becomes:\[ 2r^4 + 3r^3 - 16r^2 + 15r - 4 = 0 \]

Step 2: Solve Auxiliary Equation

Use a Computer Algebra System (CAS) to find the roots of the auxiliary equation. The solution provides the distinct roots needed to form the general solution:- \( r_1 = 2 \) (real root)- \( r_2 = -1 \) (real root)- \( r_3 = \frac{1}{2} + \frac{i\sqrt{7}}{2} \) (complex root)- \( r_4 = \frac{1}{2} - \frac{i\sqrt{7}}{2} \) (complex root)

Step 3: Form the General Solution

Using the roots from the auxiliary equation, the general solution is:\[y(t) = c_1 e^{2t} + c_2 e^{-t} + e^{\frac{1}{2}t}(c_3 \cos(\frac{\sqrt{7}}{2}t) + c_4 \sin(\frac{\sqrt{7}}{2}t))\]

Step 4: Apply Initial Conditions

Differentiate the general solution to get expressions for \(y'(t), y''(t),\) and \(y'''(t)\), then substitute \(y(0), y'(0), y''(0),\) and \(y'''(0)\) from the initial conditions:- \(y(0) = -2\)- \(y'(0) = 6\)- \(y''(0) = 3\)- \(y'''(0) = \frac{1}{2}\)

Step 5: Solve the System of Equations

Input the system of four equations derived from applying the initial conditions into the CAS:1. \( c_1 + c_2 + c_3 = -2 \)2. \( 2c_1 - c_2 + \frac{c_4 \sqrt{7}}{2} = 6 \)3. \( 4c_1 + c_2 - \frac{3}{2}c_3 = 3 \)4. \( 8c_1 - c_2 + \frac{7}{4}c_4 = \frac{1}{2} \)Find the coefficients \(c_1, c_2, c_3,\) and \(c_4\) to satisfy these equations.

Step 6: Calculate Coefficients

Using the CAS, solve the system obtained:- \(c_1 = 1\)- \(c_2 = -3\)- \(c_3 = 0\)- \(c_4 = 8\)

Key Concepts

Auxiliary Equation

The auxiliary equation is a fundamental tool in solving linear differential equations. It is formed by translating the differential equation into a polynomial equation using a special substitution: replace each derivative of the unknown function with a power of a variable, commonly denoted as \(r\), which acts as a placeholder for roots of the equation. For instance, in the given differential equation \[ 2 y^{(4)}+3 y^{\prime \prime \prime}-16 y^{\prime \prime}+15 y^{\prime}-4 y=0, \]the auxiliary equation is constructed as \[ 2r^4 + 3r^3 - 16r^2 + 15r - 4 = 0. \]Solving this equation helps to find the roots, which will be utilized to construct the general solution of the differential equation.

Initial Conditions

Initial conditions are specific values set for the solution of a differential equation at a particular point. These conditions allow us to find the unique values of integration constants in the general solution. For our exercise, they are:

• \( y(0) = -2 \)
• \( y^{\prime}(0) = 6 \)
• \( y^{\prime \prime}(0) = 3 \)
• \( y^{\prime \prime \prime}(0) = \frac{1}{2} \)

Using these initial conditions, we can generate a system of equations. Solving this system provides the coefficients needed for our particular solution. This is a critical step because it transforms the broad applicability of the general solution into a specific answer tailored to the given initial conditions.

General Solution

The general solution of a differential equation represents all possible solutions. It relies heavily on the roots found by solving the auxiliary equation. Different types of roots contribute distinct components to the overall solution:

• Real roots contribute terms of the form \( c e^{rt} \).
• Complex roots lead to terms involving sine and cosine functions.

For our exercise, the general solution is: \[ y(t) = c_1 e^{2t} + c_2 e^{-t} + e^{\frac{1}{2}t}(c_3 \cos(\frac{\sqrt{7}}{2}t) + c_4 \sin(\frac{\sqrt{7}}{2}t)) \]Here, \( c_1, c_2, c_3, \) and \( c_4 \) are constants determined by the initial conditions, ensuring that each potential solution fits exactly with the prescribed initial values.

Complex Roots

Complex roots arise when the discriminant of the quadratic factor in the auxiliary equation is negative. Each pair of complex conjugates \( \alpha \pm \beta i \) adds oscillatory components to the general solution. These components are expressed using exponential, sine, and cosine functions.The process involves creating a solution structured as: \[ e^{\alpha t} ( c_3 \cos(\beta t) + c_4 \sin(\beta t) ) \]In our scenario, the complex roots \( r_3 = \frac{1}{2} + \frac{i\sqrt{7}}{2} \) and \( r_4 = \frac{1}{2} - \frac{i\sqrt{7}}{2} \) contribute such a term. The presence of complex roots transforms the differential equation's solutions into periodic functions, reflecting the oscillating behavior of systems such as springs and electrical circuits.