Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 7: Problem 28
Obtain the general solution. $$ \left(D^{3}-3 D-2\right) y=100 \sin 2 x. $$
Short Answer
Expert verified
\(y = C_1 e^{2x} + C_2 e^{-x} + C_3 xe^{-x} + 10 \cos 2x - 25 \sin 2x\)
Step by step solution
01
Identify the Type of Differential Equation
The equation given is a linear differential equation with constant coefficients, where the left-hand side is a polynomial of the differential operator \(D\), and the right-hand side is a sinusoidal function.
02
Find the Complementary Function (CF)
Solve the homogeneous equation associated with \((D^3 - 3D - 2)y = 0\). The characteristic equation is \(r^3 - 3r - 2 = 0\). Factor or solve this polynomial to find the roots.
03
Solve the Characteristic Equation
The characteristic equation \(r^3 - 3r - 2 = 0\) factors as \((r - 2)(r^2 + 2r + 1) = 0\). Thus, the roots are \(r = 2\) and \(r = -1\) (with a repeated root).
04
Construct the Complementary Function
Given the roots \(r = 2\), \(r = -1\), \(r = -1\), the complementary function is \(y_c = C_1 e^{2x} + C_2 e^{-x} + C_3 xe^{-x}\), where \(C_1, C_2, C_3\) are constants.
05
Determine the Particular Solution (PS)
Since the non-homogeneous part is \(100 \sin 2x\), assume a particular solution of the form \(y_p = A\cos 2x + B\sin 2x\). Substitute \(y_p\) into the differential equation to solve for \(A\) and \(B\).
06
Substitute and Solve for Coefficients
After substituting \(y_p = A\cos 2x + B\sin 2x\) and its derivatives into \((D^3 - 3D - 2) y = 100 \sin 2x\), we match coefficients. Solve the resulting system of equations to find \(A = 10\) and \(B = -25\).
07
Construct the General Solution
Combine both the complementary function and the particular solution to get the general solution: \[y = C_1 e^{2x} + C_2 e^{-x} + C_3 xe^{-x} + 10 \cos 2x - 25 \sin 2x\] where \(C_1, C_2, C_3\) are arbitrary constants.
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.
Complementary Function
The complementary function (CF) is a crucial concept in the solution of linear differential equations. It's the solution to the homogeneous part of the equation. When you're dealing with an equation like \((D^3 - 3D - 2)y = 0\), you're finding the CF. The goal is to solve the equation without the non-homogeneous term (in our case, 100 \(\sin 2x\)).
Here's how you can find it:
Here's how you can find it:
- First, identify the characteristic equation by setting the left-hand side equal to zero: \(r^3 - 3r - 2 = 0\).
- Solve for \(r\) to find the roots. In this problem, the roots are \(r = 2, -1, -1\).
- Use these roots to construct the CF. Since one root is repeated, use the form: \(y_c = C_1 e^{2x} + C_2 e^{-x} + C_3 xe^{-x}\). The terms \(e^{2x}\) and \(e^{-x}\) correspond to the distinct and repeated roots.
Particular Solution
A particular solution (PS) is specifically tailored to address the non-homogeneous part of a differential equation. This is the part driven by external inputs or sources, such as the \(100 \sin 2x\) in our example.
To find the PS, target only the non-homogeneous part:
To find the PS, target only the non-homogeneous part:
- Assume a form for your PS based on the type of non-homogeneous term. For \(100 \sin 2x\), assume \(y_p = A \cos 2x + B \sin 2x\).
- Substitute this assumed form into the original differential equation. Calculate the necessary derivatives to do this.
- Compare both sides of the equation by matching coefficients. This will result in a system of equations.
- Solve these equations to find the constants \(A\) and \(B\). In this exercise, \(A\) is 10 and \(B\) is -25.
Characteristic Equation
The characteristic equation is a cornerstone concept when dealing with linear differential equations with constant coefficients. It's derived from the differential operator when you substitute \(D = r\). Let's examine its role and importance.
Here is how it fits in our exercise:
Here is how it fits in our exercise:
- The differential operator \((D^3 - 3D - 2)\) leads to the characteristic polynomial \(r^3 - 3r - 2\).
- Solving this polynomial entails finding its roots. In the provided problem, it factors into \((r - 2)(r^2 + 2r + 1) = 0\). This gives distinct roots and sometimes repeated roots.
- The roots determine the structure of the complementary function. Each root corresponds to an exponential term \(e^{rx}\) in the CF. A repeated root like \(r = -1\) results in terms like \(x e^{-x}\).
