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Magazine Chapter 17: Problem 40
Solve the differential equations.Some of the equations can be solved by the method of undetermined coefficients, but others cannot. $$y^{\prime \prime}+4 y=\sin x$$
Short Answer
Expert verified
The general solution is \( y = c_1 \cos(2x) + c_2 \sin(2x) + \frac{1}{3} \sin x \).
Step by step solution
01
Identify Type of Differential Equation
The given equation is \( y'' + 4y = \sin x \). This is a second-order linear non-homogeneous differential equation with constant coefficients.
02
Solve the Homogeneous Equation
First, solve the homogeneous equation \( y'' + 4y = 0 \). The characteristic equation is \( r^2 + 4 = 0 \), which gives roots \( r = \pm 2i \). The general solution of the homogeneous equation is \( y_h = c_1 \cos(2x) + c_2 \sin(2x) \), where \( c_1 \) and \( c_2 \) are constants.
03
Construct Particular Solution
For the non-homogeneous part, \( \sin x \), we apply the method of undetermined coefficients. Choose a particular solution of the form \( y_p = A \sin x + B \cos x \).
04
Differentiate the Particular Solution
Compute the derivatives: \( y_p' = A \cos x - B \sin x \) and \( y_p'' = -A \sin x - B \cos x \).
05
Substitute into Original Equation
Substitute \( y_p, y_p', \) and \( y_p'' \) back into the original equation: \(-A \sin x - B \cos x + 4(A \sin x + B \cos x) = \sin x\). Simplify to get \((4A - A)\sin x + (4B - B)\cos x = \sin x\).
06
Identify Unknown Coefficients
From \((3A)\sin x + (3B)\cos x = \sin x\), equate coefficients: \(3A = 1\) and \(3B = 0\). Solve these equations to find \(A = \frac{1}{3}\) and \(B = 0\). Thus, \( y_p = \frac{1}{3} \sin x \).
07
Write the General Solution
Combine the homogeneous and particular solutions: \(y = y_h + y_p = c_1 \cos(2x) + c_2 \sin(2x) + \frac{1}{3} \sin x\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Method of Undetermined Coefficients
The method of undetermined coefficients is a useful tool when dealing with non-homogeneous differential equations that have constant coefficients. It allows us to find a particular solution to the equation. The key idea is to guess a form of the particular solution based on the function on the right side of the equation, like sin, cos, exponential functions, or polynomials.
To implement this method, follow these steps:
To implement this method, follow these steps:
- Identify the type of function on the right-hand side of the differential equation (e.g., sine, cosine, exponentials).
- Propose a form for the particular solution using undetermined coefficients. For example, if you have a term like \(\sin x\), try \(y_p = A \sin x + B \cos x\).
- Substitute this proposed solution and its derivatives back into the original differential equation.
- Solve the system of equations to determine the unknown coefficients \(A\) and \(B\).
Non-homogeneous Differential Equations
Non-homogeneous differential equations contain terms that are not part of the derivative function. Unlike homogeneous equations, which equal zero when the derivative terms are summed, non-homogeneous equations have an additional function on the right side. This function is what we address when searching for a particular solution.
These equations can be modeled as:
These equations can be modeled as:
- The general form: \(a_n y^{(n)} + a_{n-1}y^{(n-1)} + \ldots + a_1 y' + a_0 y = g(x)\)
- The homogeneous part: all terms involving the derivative of \(y\)
- The non-homogeneous part: \(g(x)\), where \(g(x)\) is not zero.
Characteristic Equation
The characteristic equation is a powerful concept in solving homogeneous linear differential equations with constant coefficients. It transforms the problem of finding solutions to differential equations into solving algebraic equations.
Here's how it works:
Here's how it works:
- Write the homogeneous differential equation, such as \(y'' + py' + qy = 0\).
- Assume a solution of the form \(y = e^{rx}\). Substitute \(y\), \(y'\), and \(y''\) into the equation.
- Factor out the common \(e^{rx}\) factor (since it is never zero), leaving a polynomial in \(r\).
- Solve for \(r\) to obtain roots. These can be real or complex and lead to different forms of the general solution.
