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Magazine Chapter 19: Problem 8
Solve the differential equation by using the method of variation of parameters. $$ y^{\prime \prime}+y=\sin x $$
Short Answer
Expert verified
The general solution is \( y = C_1 \cos x + C_2 \sin x - \frac{1}{2} \cos 2x - \frac{1}{2} \).
Step by step solution
01
Solve Homogeneous Equation
To solve the given differential equation using the method of variation of parameters, we start by solving the associated homogeneous equation: \( y'' + y = 0 \).Assuming a solution of the form \( y = e^{rx} \), the characteristic equation is \( r^2 + 1 = 0 \), which gives roots \( r = \pm i \). Therefore, the general solution of the homogeneous equation is: \( y_h = C_1 \, \cos x + C_2 \, \sin x \).
02
Particular Solution Setup with Variation of Parameters
Next, we look for a particular solution \( y_p(x) \) using variation of parameters. Assume \( y_p = u_1(x) \cos x + u_2(x) \sin x \) where \( u_1(x) \) and \( u_2(x) \) are functions to be determined. The next step is to find expressions for \( u_1 \) and \( u_2 \).
03
Derivative Constraints
We impose the condition:\[ u_1'(x) \cos x + u_2'(x) \sin x = 0 \]to ensure orthogonality and simplification. Hence,\( u_1'(x) \cos x + u_2'(x) \sin x = 0 \).
04
Solve for Derivatives
Differentiate \( y_p \) to get \( y_p' = u_1'(x) \cos x - u_1(x) \sin x + u_2'(x) \sin x + u_2(x) \cos x \).Substitute \( y_p' \) into the equation for \( y'' + y = \sin x \):\[ u_1'(x) (-\sin x) + u_2'(x) \cos x = \sin x \].We now have two equations:1. \( u_1'(x) \cos x + u_2'(x) \sin x = 0 \)2. \( u_1'(x)(-\sin x) + u_2'(x) \cos x = \sin x \).
05
Solve System of Equations
Using the system of equations obtained:1. \( u_1'(x) \cos x + u_2'(x) \sin x = 0 \)2. \( u_1'(x)(-\sin x) + u_2'(x) \cos x = \sin x \),solve for \( u_1'(x) \) and \( u_2'(x) \).By substitution or Cramer's rule:\[ u_1'(x) = \sin x \quad \text{and} \quad u_2'(x) = 0 \].Integrate to find each function:\( u_1(x) = -\cos x \) and \( u_2(x) = 0 \).
06
Find Particular Solution
Substitute back to find the particular solution:\( y_p = u_1(x) \cos x + u_2(x) \sin x = (-\cos x) \cos x = \displaystyle \frac{-1}{2}(\cos 2x + 1) \).
07
General Solution
Combine the homogeneous solution and particular solution to find the general solution:\( y = y_h + y_p = C_1 \cos x + C_2 \sin x + \frac{-1}{2}(\cos 2x + 1) \).
08
Simplifying General Solution
Simplify further if needed:\( y = C_1 \cos x + C_2 \sin x - \frac{1}{2} \cos 2x - \frac{1}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equation
A differential equation involves functions and their derivatives. It represents physical phenomena such as motion, growth, and decay. In mathematics, a differential equation like \( y'' + y = \sin x \) tells us how a function \( y(x) \) behaves through its second derivative \( y'' \) and a sine wave, \( \sin x \). This particular equation is a second-order linear differential equation, indicative of the highest derivative involved (\( y'' \)), and due to its linearity, it can be solved using superposition of solutions.
Homogeneous Equation
A homogeneous differential equation is a simpler form where the right side equals zero, like \( y'' + y = 0 \). Solving this equation is key to our overall solution strategy since it represents the 'natural' behaviour of systems without external forces. To solve it, we use the characteristic polynomial, \( r^2 + 1 = 0 \), yielding complex roots \( r = \pm i \). This reflects oscillatory solutions: \( y_h = C_1 \cos x + C_2 \sin x \). These functions form the backbone of our solution, adjusted for any external influences later.
Particular Solution
The particular solution, \( y_p(x) \), manages the 'forced' part of the equation where \( y'' + y = \sin x \). The method of variation of parameters allows us to find \( y_p \) by introducing unknown functions \( u_1(x) \) and \( u_2(x) \) to modify our homogeneous solution into \( y_p = u_1(x) \cos x + u_2(x) \sin x \). By differentiating and solving based on constraints, like ensuring orthogonality, we deduce the forms of these functions to satisfy the whole equation with \( \sin x \) on the right-side. We find that \( u_1(x) = -\cos x \), while \( u_2(x) = 0 \), leading to \( y_p = \frac{-1}{2}( \cos 2x + 1 ) \).
General Solution
The general solution of a differential equation combines both homogeneous and particular solutions, covering all possible scenarios. By integrating these, we have \( y = C_1 \cos x + C_2 \sin x \), the general response, alongside \( y_p = \frac{-1}{2} \cos 2x - \frac{1}{2} \), the specific adjustment for the applied force. Together, these components form \( y = C_1 \cos x + C_2 \sin x - \frac{1}{2} \cos 2x - \frac{1}{2} \), expressing the entire behaviour of the system. This formula accounts for both innate and external dynamics, solved step-by-step using the method of variation of parameters.
