Chapter 17: Problem 27
Solve the equations by variation of parameters.
$$\frac{d^{2} y}{d x^{2}}+y=\sec x, \quad-\frac{\pi}{2}
Short Answer
Expert verified
The general solution is \( y = C_1 \cos x + C_2 \sin x + (\ln |\cos x|) \cos x + x \sin x \).
Step by step solution
01
Solve the Homogeneous Equation
First, solve the homogeneous equation associated with the given differential equation. The homogeneous equation is \( \frac{d^{2} y}{d x^{2}} + y = 0 \). Assume a solution of the form \( y = e^{rx} \). Then the characteristic equation is \( r^2 + 1 = 0 \). This gives characteristic roots \( r = i \) and \( r = -i \), leading to the general solution \( y_h = C_1 \cos x + C_2 \sin x \).
02
Identify Particular Solution using Variation of Parameters
Given that the solutions to the complementary equation \( y_h = C_1 \cos x + C_2 \sin x \), we use variation of parameters. Assume a particular solution of the form \( y_p = u_1(x) \cos x + u_2(x) \sin x \) where \( u_1(x) \) and \( u_2(x) \) are functions to be determined.
03
Determine Functions u_1 and u_2
To find \( u_1(x) \) and \( u_2(x) \), use the formulas: \[u_1^\prime = - \frac{\sin x \cdot \, \sec x}{\cos x \cdot \sin x' - \sin x \cdot \cos x'} = -\tan x\]\[u_2^\prime = \frac{\cos x \cdot \, \sec x}{\cos x \cdot \sin x' - \sin x \cdot \cos x'} = 1 \]Here, the Wronskian is 1 since \( \cos^2 x + \sin^2 x = 1 \). Integrate \( u_1 \)' and \( u_2 \)' to find \( u_1(x) \) and \( u_2(x) \):\[ u_1(x) = \int -\tan x \, dx = \ln |\cos x| \]\[ u_2(x) = \int 1 \, dx = x \]
04
Construct the Particular Solution
Combine \( u_1(x) \) and \( u_2(x) \) into the particular solution:\[ y_p = (\ln |\cos x|) \cos x + x \sin x. \]
05
Write the General Solution
The general solution of the differential equation is the sum of the homogeneous and particular solutions:\[ y = y_h + y_p = C_1 \cos x + C_2 \sin x + (\ln |\cos x|) \cos x + x \sin x. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are mathematical equations that involve derivatives of a function. They help describe various phenomena where the rate of change is crucial, such as physics, engineering, and other scientific fields. In our problem, we are dealing with a second-order differential equation, which includes the second derivative of the function \( y \) with respect to \( x \). This equation is \( \frac{d^{2} y}{d x^{2}} + y = \sec x \).
This equation is considered non-homogeneous because it is not equal to zero. The presence of the \( \sec x \) term makes it non-homogeneous, as opposed to homogeneous equations where there would be no term on the right-hand side. The goal here is to find a particular solution for this non-homogeneous equation.
This equation is considered non-homogeneous because it is not equal to zero. The presence of the \( \sec x \) term makes it non-homogeneous, as opposed to homogeneous equations where there would be no term on the right-hand side. The goal here is to find a particular solution for this non-homogeneous equation.
Homogeneous Equation
A homogeneous equation is a type of differential equation where all its terms are homogeneous functions with the same degree. For our task, the homogeneous version of the original equation is \( \frac{d^{2} y}{d x^{2}} + y = 0 \).
This means there are no external functions like \( \sec x \) affecting it. We use the characteristic equation to solve it, assuming solutions of the form \( y = e^{rx} \).
Solving gives us the roots \( r = i \) and \( r = -i \), which are imaginary. This leads to a general homogeneous solution \( y_h = C_1 \cos x + C_2 \sin x \), utilizing Euler's formula for representing complex exponentials.
This means there are no external functions like \( \sec x \) affecting it. We use the characteristic equation to solve it, assuming solutions of the form \( y = e^{rx} \).
Solving gives us the roots \( r = i \) and \( r = -i \), which are imaginary. This leads to a general homogeneous solution \( y_h = C_1 \cos x + C_2 \sin x \), utilizing Euler's formula for representing complex exponentials.
Particular Solution
In the realm of differential equations, a particular solution refers to a solution that satisfies the non-homogeneous equation. To find it, we use a method known as variation of parameters.
For our problem, given that we already have the complementary solution \( y_h = C_1 \cos x + C_2 \sin x \), we assume a particular solution of the form \( y_p = u_1(x) \cos x + u_2(x) \sin x \), where \( u_1(x) \) and \( u_2(x) \) are unknown functions.
By determining these functions using integration, we find \( u_1(x) = \ln |\cos x| \) and \( u_2(x) = x \). These lead us to the particular solution \( y_p = (\ln |\cos x|) \cos x + x \sin x \). This fits the non-homogeneous differential equation by catering for the additional \( \sec x \) term.
For our problem, given that we already have the complementary solution \( y_h = C_1 \cos x + C_2 \sin x \), we assume a particular solution of the form \( y_p = u_1(x) \cos x + u_2(x) \sin x \), where \( u_1(x) \) and \( u_2(x) \) are unknown functions.
By determining these functions using integration, we find \( u_1(x) = \ln |\cos x| \) and \( u_2(x) = x \). These lead us to the particular solution \( y_p = (\ln |\cos x|) \cos x + x \sin x \). This fits the non-homogeneous differential equation by catering for the additional \( \sec x \) term.
Characteristic Roots
Characteristic roots are crucial in solving homogeneous differential equations. These roots are derived from the characteristic equation, formed by assuming a potential solution of the form \( y = e^{rx} \).
For the homogeneous equation \( \frac{d^{2} y}{d x^{2}} + y = 0 \), its characteristic equation is \( r^2 + 1 = 0 \). Solving it gives us the roots \( r = i \) and \( r = -i \). The roots being imaginary indicates sinusoidal components in the solution.
Hence, the general solution is expressed as \( y_h = C_1 \cos x + C_2 \sin x \), which is a linear combination of sine and cosine functions. These roots help understand the oscillatory nature of the system described by the differential equation.
For the homogeneous equation \( \frac{d^{2} y}{d x^{2}} + y = 0 \), its characteristic equation is \( r^2 + 1 = 0 \). Solving it gives us the roots \( r = i \) and \( r = -i \). The roots being imaginary indicates sinusoidal components in the solution.
Hence, the general solution is expressed as \( y_h = C_1 \cos x + C_2 \sin x \), which is a linear combination of sine and cosine functions. These roots help understand the oscillatory nature of the system described by the differential equation.