Chapter 10: Problem 3
Find the general solution to the differential equation using variation of parameters. $$y^{\prime \prime}+4 y=\sec x$$
Short Answer
Expert verified
The general solution is \( y = C_1 \cos(2x) + C_2 \sin(2x) - \frac{1}{4}\ln|\sec x + \tan x| \cos(2x) + \frac{1}{4}\ln|\sec x + \tan x| \sin(2x) \).
Step by step solution
01
Solve the Homogeneous Equation
The given differential equation is \( y'' + 4y = \sec x \). First, solve the homogeneous equation \( y'' + 4y = 0 \). The characteristic equation is \( r^2 + 4 = 0 \), which simplifies to \( r^2 = -4 \). Solving for \( r \), we get \( r = \pm 2i \). The general solution of the homogeneous equation is \( y_h = C_1 \cos(2x) + C_2 \sin(2x) \).
02
Set Up Variation of Parameters
To find a particular solution using variation of parameters, we assume a solution of the form \( y = u_1(x) \cos(2x) + u_2(x) \sin(2x) \), where \( u_1(x) \) and \( u_2(x) \) are functions to be determined. The formulas for \( u_1'(x) \) and \( u_2'(x) \) are derived from\[ u_1' = -\frac{y_2 \cdot g(x)}{W(y_1, y_2)}\]\[ u_2' = \frac{y_1 \cdot g(x)}{W(y_1, y_2)}\]where \( y_1 = \cos(2x) \), \( y_2 = \sin(2x) \), and \( g(x) = \sec x \).
03
Calculate the Wronskian
The Wronskian of \( y_1 \) and \( y_2 \) is calculated as follows:\[W(y_1, y_2) = \det\begin{vmatrix} \cos(2x) & \sin(2x) \ -2\sin(2x) & 2\cos(2x) \end{vmatrix} = 2(\cos^2(2x) + \sin^2(2x)) = 2\]Thus, the Wronskian \( W = 2 \).
04
Integrate to Find \( u_1(x) \) and \( u_2(x) \)
Using the formulas from Step 2 and the Wronskian from Step 3, we find:\[u_1' = -\frac{\sin(2x) \sec x}{2} = -\frac{\sin(2x)}{2\cos x}\]\[u_2' = \frac{\cos(2x) \sec x}{2} = \frac{\cos(2x)}{2\cos x}\]Now, integrate these expressions:\[u_1 = \int -\frac{\sin(2x)}{2\cos x} \, dx \]\[u_2 = \int \frac{\cos(2x)}{2\cos x} \, dx \]By simplifying and integrating, we find:\[u_1 = -\frac{1}{4}\ln|\sec x + \tan x| + C_1\]\[u_2 = \frac{1}{4}\ln|\sec x + \tan x| + C_2\]
05
Form the Particular Solution
Using \( u_1(x) \) and \( u_2(x) \) derived in the previous step, the particular solution \( y_p \) is:\[y_p = u_1(x) \cos(2x) + u_2(x) \sin(2x)\]Substitute the expressions for \( u_1(x) \) and \( u_2(x) \) to give:\[y_p = \left(-\frac{1}{4}\ln|\sec x + \tan x| + C_1\right) \cos(2x) + \left(\frac{1}{4}\ln|\sec x + \tan x| + C_2\right) \sin(2x)\]
06
Write the General Solution
The general solution \( y \) of the differential equation is the sum of the homogeneous solution \( y_h \) and the particular solution \( y_p \):\[y = y_h + y_p \]\[y = C_1 \cos(2x) + C_2 \sin(2x) - \frac{1}{4}\ln|\sec x + \tan x| \cos(2x) + \frac{1}{4}\ln|\sec x + \tan x| \sin(2x)\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
A differential equation involves derivatives of a function and expresses how a rate of change depends on the variables involved. In our example, the differential equation is given as \( y'' + 4y = \sec x \). This means we are working with a second-order differential equation where derivatives up to the second order are present. The right side, \( \sec x \), is the non-homogeneous part of the equation, which shows a direct relationship between the function's derivatives and the input variable \( x \). Differential equations can describe various real-world phenomena such as motion, heat flow, or even population dynamics. Understanding the structure allows one to predict and analyze system behaviors.
Homogeneous Equation
A homogeneous equation results when there's no external force or input, denoted by setting the right side to zero. In this case, the homogeneous version is \( y'' + 4y = 0 \). Solving it helps in finding solutions that only depend on internal system dynamics. By exploring the characteristic equation, which is obtained by replacing derivatives with algebraic variables, \( r^2 + 4 = 0 \), we derive the roots \( r = \pm 2i \). These complex roots are related to oscillating solutions. The general solution for this equation is a combination of sine and cosine functions: \( y_h = C_1 \cos(2x) + C_2 \sin(2x) \). These solutions form the basis for finding particular solutions when dealing with non-homogeneous parts.
Wronskian
The Wronskian is a determinant used to check if a set of solutions is linearly independent. For two functions, such as \( y_1 = \cos(2x) \) and \( y_2 = \sin(2x) \), the Wronskian is determined like this: \[W(y_1, y_2) = \begin{vmatrix} \cos(2x) & \sin(2x) \ -2\sin(2x) & 2\cos(2x) \end{vmatrix}\]Calculating, we find:\[W = 2(\cos^2(2x) + \sin^2(2x)) = 2\]A non-zero Wronskian confirms that the functions are linearly independent, making them valid components in constructing a solution for the differential equation.
Particular Solution
A particular solution is a specific instance of a solution to a non-homogeneous differential equation. In our method, variation of parameters, we assume the general solution involves functions \( u_1(x) \) and \( u_2(x) \) combined with known solutions \( \cos(2x) \) and \( \sin(2x) \). By finding derivatives like \( u_1' \) and \( u_2' \), we set up integrals:\[ u_1 = \int -\frac{\sin(2x)}{2\cos x} \, dx \]\[ u_2 = \int \frac{\cos(2x)}{2\cos x} \, dx \]These integrated forms yield expressions including logarithms, such as \(-\frac{1}{4}\ln|\sec x + \tan x|\). The complete particular solution is then formed by combining these results with \( y_1 \) and \( y_2 \), resulting in an expression that fits the original differential equation including the forcing function \( \sec x \). The complete solution for the equation combines this particular solution with the homogeneous solution to fully solve the differential equation.