Chapter 7: Problem 23
Find the particular solution indicated. \(\left(D^{3}-3 D-2\right) y=0 ;\) when \(x=0, y=0, y^{\prime}=9, y^{\prime \prime}=0\).
Short Answer
Expert verified
The particular solution is \(y(x) = 2e^{2x} - 2e^{-x} - 3xe^{-x}\).
Step by step solution
01
Identify the Differential Operator
The given differential equation is \((D^3 - 3D - 2)y = 0\). This indicates a third-order linear homogeneous differential equation.
02
Find the Auxiliary Equation
The auxiliary equation is derived by replacing each derivative operator \(D\) with \(r\), giving us \(r^3 - 3r - 2 = 0\).
03
Solve for the Roots of the Auxiliary Equation
To solve \(r^3 - 3r - 2 = 0\), we try potential rational roots using the Rational Root Theorem and find \(r = 2\) works, then factor the equation to \((r - 2)(r^2 + 2r + 1) = 0\). The quadratic \(r^2 + 2r + 1 = 0\) can be factored further as \((r + 1)^2 = 0\). Thus, the roots are \(r = 2\) and \(r = -1\) with multiplicity 2.
04
Construct the General Solution
The general solution, based on the roots, is \(y(x) = C_1 e^{2x} + C_2 e^{-x} + C_3 x e^{-x}\) as \(r = -1\) has multiplicity 2.
05
Apply Initial Conditions
Substitute the initial conditions: \(y(0) = 0\), \(y'(0) = 9\), \(y''(0) = 0\) into the general solution to find the constants.\- First condition: \(C_1 + C_2 = 0\).\- Differentiate the general solution for \(y'(x)\): \(y'(x) = 2C_1 e^{2x} - C_2 e^{-x} - C_3 e^{-x} + C_3 x e^{-x}\). Substituting \(x=0\), we have \(2C_1 - C_2 - C_3 = 9\).\- Differentiating again gives \(y''(x) = 4C_1 e^{2x} + C_2 e^{-x} + 2C_3 e^{-x} - C_3 x e^{-x}\), substituting \(x=0\), we obtain \(4C_1 + C_2 + 2C_3 = 0\).
06
Solve the System of Equations for Constants
Solve \(C_1 + C_2 = 0\), \(2C_1 - C_2 - C_3 = 9\), \(4C_1 + C_2 + 2C_3 = 0\).\- From \(C_1 + C_2 = 0\), we have \(C_2 = -C_1\).\- Substitute \(C_2 = -C_1\) into the other equations gives:\1. \(2C_1 + C_1 - C_3 = 9\) simplifying to \(3C_1 - C_3 = 9\).\2. \(4C_1 - C_1 + 2C_3 = 0\) simplifying to \(3C_1 + 2C_3 = 0\).\- Solve the simplified system: from \(3C_1 + 2C_3 = 0\), \(C_3 = -\frac{3}{2} C_1\). Substitute into \(3C_1 - C_3 = 9\) gives \(3C_1 + \frac{3}{2} C_1 = 9\), yielding \(C_1 = 2\).\- Therefore, \(C_2 = -2\), \(C_3 = -3\).
07
Write the Particular Solution
Substitute the constants back into the general solution to find the particular solution: \(y(x) = 2 e^{2x} - 2 e^{-x} - 3x e^{-x}\).
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Differential Equations
Homogeneous differential equations are a fundamental concept in mathematics, specifically in calculus and differential equations. These equations are called "homogeneous" because their format ensures that any term without a derivative is zero. In simpler terms, all terms involve derivatives of the unknown function, often set to zero. For instance, the given equation \( (D^3 - 3D - 2) y = 0 \) is a third-order linear homogeneous differential equation.
Such equations generally appear in the format \( Ly = 0 \), where \( L \) is a linear differential operator. In this case, \( L = D^3 - 3D - 2 \). These equations are important in modeling natural phenomena where the rate of change of a variable is related to its current state without external influences.
Such equations generally appear in the format \( Ly = 0 \), where \( L \) is a linear differential operator. In this case, \( L = D^3 - 3D - 2 \). These equations are important in modeling natural phenomena where the rate of change of a variable is related to its current state without external influences.
- They often help describe systems in equilibrium.
- Understanding their solutions is key to studying how systems evolve over time.
Initial Conditions
Initial conditions are crucial in finding a specific solution, or particular solution, to a differential equation. While a general solution solves the differential equation universally, initial conditions narrow this down to a unique path.
In our problem, the initial conditions are \( y(0) = 0 \), \( y'(0) = 9 \), and \( y''(0) = 0 \). These conditions specify values of the function and its derivatives at \( x = 0 \), guiding us in fine-tuning the arbitrary constants in the general solution.
In our problem, the initial conditions are \( y(0) = 0 \), \( y'(0) = 9 \), and \( y''(0) = 0 \). These conditions specify values of the function and its derivatives at \( x = 0 \), guiding us in fine-tuning the arbitrary constants in the general solution.
- Initial conditions transform the general solution into a particular solution.
- The solutions must satisfy these conditions at the given point.
Auxiliary Equation
The auxiliary equation is a vital tool in solving linear differential equations, especially for homogeneous types. This equation arises by substituting each derivative operator \( D \) in the original differential equation with \( r \,\text{(a real or complex number)}\), transforming the problem into a polynomial equation.
In our specific example, starting from the differential equation \( (D^3 - 3D - 2) y = 0 \,\) we construct the auxiliary equation \( r^3 - 3r - 2 = 0 \.\).
In our specific example, starting from the differential equation \( (D^3 - 3D - 2) y = 0 \,\) we construct the auxiliary equation \( r^3 - 3r - 2 = 0 \.\).
- This step simplifies the problem, allowing us to find possible solutions for \( y \).
- We solve for \( r \) to find characteristic roots, which lead us to the general solution format.
General Solution
The general solution for a homogeneous differential equation encapsulates all potential solutions to that equation without initial condition constraints. For the equation \( (D^3 - 3D - 2) y = 0 \,\) solving the auxiliary equation \( r^3 - 3r - 2 = 0 \), we find roots \( r = 2 \) and \( r = -1 \) with multiplicity 2.
Using these roots, the general solution is constructed as:
\[y(x) = C_1 e^{2x} + C_2 e^{-x} + C_3 x e^{-x}\\]
This formula represents the general solution.
Using these roots, the general solution is constructed as:
\[y(x) = C_1 e^{2x} + C_2 e^{-x} + C_3 x e^{-x}\\]
This formula represents the general solution.
- Each term in the expression corresponds to a root of the auxiliary equation.
- The constant coefficients \( C_1, C_2, \) and \( C_3 \) provide flexibility in shaping solutions.