Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 8: Problem 32
Obtain the general solution. $$ y^{\prime \prime}+6 y^{\prime}+13 y=60 \cos x+26 $$
Short Answer
Expert verified
The general solution is \( y = e^{-3x}(C_1 \cos(2x) + C_2 \sin(2x)) + 5\cos x + 2 \).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is a non-homogeneous linear second-order differential equation: \( y'' + 6y' + 13y = 60\cos x + 26 \).
02
Solve the Associated Homogeneous Equation
First, solve the homogeneous equation \( y'' + 6y' + 13y = 0 \). Find the characteristic equation: \( r^2 + 6r + 13 = 0 \).
03
Determine Characteristic Roots
Solve the characteristic equation \( r^2 + 6r + 13 = 0 \) using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 6, c = 13 \). This gives \( r = \frac{-6 \pm \sqrt{-16}}{2} = -3 \pm 2i \).
04
Write the General Solution to the Homogeneous Equation
Since the roots are complex \( -3 \pm 2i \), the general solution to the homogeneous equation is \( y_h = e^{-3x}(C_1 \cos(2x) + C_2 \sin(2x)) \).
05
Find a Particular Solution to the Non-Homogeneous Equation
Guess a particular solution of form \( y_p = A\cos x + B \) due to the right-hand side. Substitute \( y_p' \) and \( y_p'' \) into the original equation: \( -A \cos x - 6A \sin x + 13A \cos x + 13B = 60\cos x + 26 \).
06
Solve for Coefficients in the Particular Solution
Equate coefficients: From \( (13A - A) \cos x + (-6A) \sin x + 13B = 60\cos x + 26 \) you get \( 12A = 60 \) and \( -6A = 0 \), so \( A = 5 \) and \( B = 2 \). Hence, \( y_p = 5\cos x + 2 \).
07
Combine Homogeneous and Particular Solutions
The general solution to the differential equation is given by combining the homogeneous and particular solutions: \( y = y_h + y_p = e^{-3x}(C_1 \cos(2x) + C_2 \sin(2x)) + 5\cos x + 2 \).
08
General Solution
The overall general solution to the differential equation is \( y = e^{-3x}(C_1 \cos(2x) + C_2 \sin(2x)) + 5\cos x + 2 \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Equation
In differential equations, a homogeneous equation is one where all terms are dependent on the function and its derivatives, without any external input. The structure of such equations can be represented as:
- The dependent variable is set to 0 on one side.
- There is no constant or function of any variables that doesn鈥檛 involve the dependent variable or its derivatives.
Non-Homogeneous Equation
A non-homogeneous equation differs in that it includes an additional function or constant on the other side of the equation apart from the function and its derivatives. Essentially, this means:
- There is an external input or forcing function, such as \( 60\cos x + 26 \) in our case.
- Solving involves both the homogeneous solution and a particular solution that corresponds to the non-homogeneous part.
Characteristic Equation
The characteristic equation is a crucial step in solving linear differential equations, particularly those that are homogeneous. To derive it, we assume a solution of the form \( y = e^{rt} \). This translates our differential equation into an algebraic equation in terms of \( r \), known as the characteristic equation. Solving it provides:
- The roots, \( r \), suggest the type of exponential and trigonometric functions that form the solution.
- In our example, \( r^2 + 6r + 13 = 0 \), leading to complex roots \( -3 \pm 2i \) when solved using the quadratic formula.
Quadratic Formula
The quadratic formula is an essential tool for solving quadratic equations that arise during the solution of differential equations. In general, a quadratic equation is of the form \( ax^2 + bx + c = 0 \). The roots are given by:
- \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- The discriminant \( b^2 - 4ac \) indicates the nature of the roots (real or complex).
