Chapter 17: Problem 24
Solve the equations by variation of parameters. $$y^{\prime \prime}-y=\sin x$$
Short Answer
Expert verified
Solve the homogeneous equation, find a particular solution using variation of parameters, and write the general solution.
Step by step solution
01
Identify the Differential Equation
The given differential equation is a second-order linear non-homogeneous equation of the form \( y'' - y = \sin x \).
02
Solve the Homogeneous Equation
To find the complementary solution \( y_c \), solve the associated homogeneous equation \( y'' - y = 0 \). The characteristic equation is \( r^2 - 1 = 0 \).
03
Find the Roots of Characteristic Equation
The characteristic equation \( r^2 - 1 = 0 \) gives roots \( r = 1 \) and \( r = -1 \).
04
Write the Complementary Solution
The complementary solution is \( y_c = C_1 e^x + C_2 e^{-x} \), where \( C_1 \) and \( C_2 \) are constants.
05
Determine Particular Solution Using Variation of Parameters
To find a particular solution \( y_p \), use the variation of parameters method. Assume \( y_p = u_1 e^x + u_2 e^{-x} \).
06
Compute Wronskian
Calculate the Wronskian \( W \) of the solutions \( e^x \) and \( e^{-x} \): \( W = e^x \cdot e^{-x} - e^{-x} \cdot (-e^x) = 2 \).
07
Set Up Integrals for Variation of Parameters
Use the formulas \( u_1' = \frac{-e^{-x} \sin x}{W} \) and \( u_2' = \frac{e^x \sin x}{W} \) to find \( u_1 \) and \( u_2 \). Since \( W = 2 \), this simplifies to \( u_1' = \frac{-e^{-x} \sin x}{2} \) and \( u_2' = \frac{e^x \sin x}{2} \).
08
Integrate to Find \( u_1 \) and \( u_2 \)
Calculate \( u_1 = \int \frac{-e^{-x} \sin x}{2} \, dx \) and \( u_2 = \int \frac{e^x \sin x}{2} \, dx \). Using integration by parts, solve these integrals.
09
Substitute to Find Particular Solution
After finding \( u_1 \) and \( u_2 \), substitute back into \( y_p = u_1 e^x + u_2 e^{-x} \) to obtain the particular solution.
10
Write the General Solution
The general solution is \( y = y_c + y_p = C_1 e^x + C_2 e^{-x} + y_p \), where \( y_p \) is the particular solution found.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Variation of Parameters
Variation of Parameters is a classical method used in solving non-homogeneous linear differential equations. This technique helps find a particular solution to equations that cannot be solved simply by initial complementary solutions. This method is particularly helpful because:
- It extends the solution space to accommodate non-homogeneous parts like external forces or inputs, such as \( \sin x \) in our case.
- It is adaptable and can be used when other methods, like undetermined coefficients, are not applicable.
Complementary Solution
The complementary solution, often denoted as \( y_c \), is the solution to the associated homogeneous differential equation. In simpler terms, when given a differential equation like \( y'' - y = \sin x \), the complementary solution comes from the homogeneous part, \( y'' - y = 0 \).
- First, we convert the differential equation to a characteristic equation: for \( y'' - y = 0 \), it is \( r^2 - 1 = 0 \).
- Next, solve the characteristic equation to find the roots. Here, the roots are \( r = 1 \) and \( r = -1 \).
- The solution then becomes a combination of exponential functions: \( y_c = C_1 e^x + C_2 e^{-x} \).
Wronskian
The Wronskian is a determinant used to determine the linear independence of functions, which is crucial for confirming that our chosen solutions \( e^x \) and \( e^{-x} \) are linearly independent in the complementary solution. In the context of this exercise:
- The Wronskian \( W \) for functions \( e^x \) and \( e^{-x} \) is calculated using the formula \( W = e^x \cdot e^{-x} - e^{-x} \cdot (-e^x) \).
- Substitute the terms to get \( W = e^{0} + e^{0} = 2 \).
Integration by Parts
Integration by Parts is a mathematical technique often used to solve integrals involving products of functions. In the context of differential equations, it's used to solve integrals like \( \int \frac{-e^{-x} \sin x}{2} \, dx \) and \( \int \frac{e^x \sin x}{2} \, dx \) to find \( u_1 \) and \( u_2 \).This technique is based on the integration formula:\[\int u \, dv = uv - \int v \, du\]where \( u \) and \( dv \) are parts of the integrand. Applying integration by parts involves:
- Choosing which parts of the integrand will be \( u \) and \( dv \).
- Deriving \( v \) from \( dv \) and \( du \) from \( u \), then substituting into the formula.
- Solving the resulting integrals to find the solutions \( u_1 \) and \( u_2 \).