Chapter 31: Problem 24
Solve the given differential equations. $$D^{2} y-D y+y=x+\sin x$$
Short Answer
Expert verified
General solution: \( y = e^{\frac{1}{2}x}(C_1 \cos(\frac{\sqrt{3}}{2}x) + C_2 \sin(\frac{\sqrt{3}}{2}x)) + x + \frac{1}{2} \sin x + \frac{1}{2} \cos x \).
Step by step solution
01
Identify the Type of Differential Equation
The given equation is a linear non-homogeneous second-order differential equation, represented as \( D^2 y - Dy + y = x + \sin x \).
02
Write the Characteristic Equation
For the homogeneous part \( D^2 y - Dy + y = 0 \), we set up the characteristic equation: \( m^2 - m + 1 = 0 \). We find the roots of the characteristic equation to solve the homogeneous part.
03
Find the Roots of the Characteristic Equation
To solve \( m^2 - m + 1 = 0 \), we apply the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1, b = -1, c = 1 \). Thus, \( m = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2} \).
04
Write the General Solution for the Homogeneous Part
The solution to the homogeneous equation is \( y_h = e^{\frac{1}{2}x}(C_1 \cos \left( \frac{\sqrt{3}}{2}x \right) + C_2 \sin \left( \frac{\sqrt{3}}{2}x \right)) \), where \( C_1 \) and \( C_2 \) are constants.
05
Determine a Particular Solution
Since the non-homogeneous part is \( x + \sin x \), we use the method of undetermined coefficients. Assume \( y_p = Ax + B + C\sin x + D\cos x \). Substitute \( y_p \) into the differential equation and equate coefficients to find \( A, B, C, \) and \( D \).
06
Substitute and Solve for Coefficients
Calculate \( D y_p = A + C \cos x - D \sin x \) and \( D^2 y_p = -C \sin x - D \cos x \). Substitute these into \( D^2 y - D y + y = x + \sin x \) and equate to find \( A = 1, B = 0, C = \frac{1}{2}, D = \frac{1}{2} \).
07
Write the Particular Solution
With the determined coefficients, the particular solution is \( y_p = x + \frac{1}{2} \sin x + \frac{1}{2} \cos x \).
08
Form the General Solution
The general solution is the sum of the homogeneous and particular solutions: \( y = e^{\frac{1}{2}x}(C_1 \cos \left( \frac{\sqrt{3}}{2}x \right) + C_2 \sin \left( \frac{\sqrt{3}}{2}x \right)) + x + \frac{1}{2} \sin x + \frac{1}{2} \cos x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Non-Homogeneous Differential Equation
In mathematics, a differential equation attempts to describe relationships involving functions and their derivatives. When a second-order linear differential equation contains terms that are not dependent on the unknown function, it is considered non-homogeneous. An example of such an equation is the given equation: \( D^2 y - Dy + y = x + \sin x \).
This equation combines derivatives of \( y \) with respect to \( x \), along with non-zero terms like \( x \) and \( \sin x \), which makes it non-homogeneous.
This equation combines derivatives of \( y \) with respect to \( x \), along with non-zero terms like \( x \) and \( \sin x \), which makes it non-homogeneous.
- "Linear" indicates that both the function and its derivatives are only multiplied by constants or zero-order functions.
- "Non-homogeneous" means that as opposed to a homogeneous equation where everything would sum to zero, a piece of the equation is influenced by external factors, such as specific functions.
Characteristic Equation
To find the solution to a linear differential equation, especially a non-homogeneous one, it helps to first solve the associated homogeneous equation. This is accomplished by determining its characteristic equation. For the equation \( D^2 y - Dy + y = 0 \), the characteristic equation helps us discover the roots related to the solutions of the homogeneous part.
For this problem, the characteristic equation is represented as \( m^2 - m + 1 = 0 \). The roots of this equation can indicate the form of the complementary or homogeneous solution.
For this problem, the characteristic equation is represented as \( m^2 - m + 1 = 0 \). The roots of this equation can indicate the form of the complementary or homogeneous solution.
- This equation is derived by substituting \( m \) for the derivative operator \( D \).
- The quadratic format makes it possible to apply the quadratic formula: \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
Method of Undetermined Coefficients
With the characteristic equation established, tackling the non-homogeneous aspect requires an effective approach like the method of undetermined coefficients. This method finds a specific, or particular solution to the non-homogeneous part of the equation.
In our case, where the additional terms are simple functions like \( x \) and \( \sin x \), we assume a particular solution using a form like \( y_p = Ax + B + C\sin x + D\cos x \). By substituting this equation into the original differential equation:
In our case, where the additional terms are simple functions like \( x \) and \( \sin x \), we assume a particular solution using a form like \( y_p = Ax + B + C\sin x + D\cos x \). By substituting this equation into the original differential equation:
- Equate derived coefficients from both sides.
- Solve for the unknown coefficients like \( A, B, C, \) and \( D \).
General Solution of Differential Equations
The goal in solving a differential equation is to determine the general solution which represents all possible solutions. For non-homogeneous differential equations, this involves summing an array of solutions.
The general solution is expressed as:
\[ y = y_h + y_p \]
where:
Consequently, it translates into a complete, comprehensive expression covering all potentialities described by the initial differential equation.
The general solution is expressed as:
\[ y = y_h + y_p \]
where:
- \( y_h \) is the complementary solution or the solution to the associated homogeneous equation.
- \( y_p \) is the particular solution obtained for the non-homogeneous equation using methods like undetermined coefficients.
Consequently, it translates into a complete, comprehensive expression covering all potentialities described by the initial differential equation.