Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 6: Problem 5
Find the solution of the differential equation that satisfies the given boundary condition\((s)\) $$f^{\prime \prime}-f^{\prime}-f=0, f(0)=0, f(1)=1$$
Short Answer
Expert verified
The solution is \( f(x) = \frac{1}{A} \left(e^{(1+\sqrt{5})/2 \cdot x} - e^{(1-\sqrt{5})/2 \cdot x}\right) \), where \( A = e^{(1+\sqrt{5})/2} - e^{(1-\sqrt{5})/2} \).
Step by step solution
01
Identify the Type of Differential Equation
The equation given is a second-order linear homogeneous differential equation of the form: \( f'' - f' - f = 0 \). We need to find a solution that satisfies this equation along with the boundary conditions.
02
Find the Characteristic Equation
To solve the differential equation, we first find the characteristic equation by assuming a solution of the form \( f = e^{rx} \). Substituting \( f = e^{rx} \), \( f' = re^{rx} \), and \( f'' = r^2 e^{rx} \) into the differential equation gives: \[ r^2 e^{rx} - r e^{rx} - e^{rx} = 0 \]Dividing the entire equation by \( e^{rx} \) (assuming \( e^{rx} eq 0 \), which is valid), we get the characteristic equation:\[ r^2 - r - 1 = 0 \]
03
Solve the Characteristic Equation
The characteristic equation is \( r^2 - r - 1 = 0 \). Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -1 \), we have:\[ r = \frac{1 \pm \sqrt{5}}{2} \]Thus, the roots are \( r_1 = \frac{1 + \sqrt{5}}{2} \) and \( r_2 = \frac{1 - \sqrt{5}}{2} \).
04
Write the General Solution
The general solution to the differential equation is given by a linear combination of the solutions corresponding to the roots of the characteristic equation:\[ f(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \]Substituting the roots, we have:\[ f(x) = C_1 e^{(1+\sqrt{5})/2 \cdot x} + C_2 e^{(1-\sqrt{5})/2 \cdot x} \]
05
Apply Boundary Conditions
We now apply the boundary conditions \( f(0) = 0 \) and \( f(1) = 1 \) to determine \( C_1 \) and \( C_2 \).- Applying \( f(0) = 0 \):\[ 0 = C_1 e^0 + C_2 e^0 = C_1 + C_2 \]This simplifies to \( C_1 + C_2 = 0 \).- Applying \( f(1) = 1 \):\[ 1 = C_1 e^{(1+\sqrt{5})/2} + C_2 e^{(1-\sqrt{5})/2} \]
06
Solve for Constants
From \( C_1 + C_2 = 0 \), we know \( C_2 = -C_1 \). Substituting this into the second boundary condition:\[ 1 = C_1\left(e^{(1+\sqrt{5})/2} - e^{(1-\sqrt{5})/2}\right) \]Let \( A = e^{(1+\sqrt{5})/2} - e^{(1-\sqrt{5})/2} \), then \( C_1 = \frac{1}{A} \) and \( C_2 = -\frac{1}{A} \).Thus, \( C_1 \) and \( C_2 \) are given by \( C_1 = \frac{1}{A} \) and \( C_2 = -\frac{1}{A} \).
07
Write the Specific Solution
Finally, substituting for \( C_1 \) and \( C_2 \), the function \( f(x) \) becomes:\[ f(x) = \frac{1}{A} \left(e^{(1+\sqrt{5})/2 \cdot x} - e^{(1-\sqrt{5})/2 \cdot x}\right) \]This is the solution to the differential equation with the given boundary conditions.
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.
Characteristic Equation
A characteristic equation is a crucial part of solving linear differential equations, especially second-order ones. When faced with a homogeneous differential equation like this one: \( f'' - f' - f = 0 \), solving it involves assuming a solution of the form \( f = e^{rx} \).
This assumption transforms the differential equation into an algebraic equation by substituting derivatives: \( f' = re^{rx} \) and \( f'' = r^2 e^{rx} \). By replacing these into the original differential equation, we reach:
This assumption transforms the differential equation into an algebraic equation by substituting derivatives: \( f' = re^{rx} \) and \( f'' = r^2 e^{rx} \). By replacing these into the original differential equation, we reach:
- \( r^2 e^{rx} - r e^{rx} - e^{rx} = 0 \)
- Dividing through by \( e^{rx} \) gives the characteristic equation: \( r^2 - r - 1 = 0 \)
Boundary Conditions
Boundary conditions are additional constraints that a solution to a differential equation must satisfy. They are necessary for finding a specific solution from the general solution of a differential equation.
In this problem, we have two boundary conditions: \( f(0) = 0 \) and \( f(1) = 1 \). Let's break down their role:
In this problem, we have two boundary conditions: \( f(0) = 0 \) and \( f(1) = 1 \). Let's break down their role:
- The boundary condition \( f(0) = 0 \) helps establish a relationship between constants in the general solution. If the general solution is \( f(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \), substituting \( x = 0 \) simplifies this to \( C_1 + C_2 = 0 \).
- The second condition \( f(1) = 1 \) allows us to identify the actual values of these constants, given the derived relationship. By substituting \( x = 1 \), it sets up another equation that, together with the first condition, can be solved for \( C_1 \) and \( C_2 \).
Homogeneous Differential Equation
A homogeneous differential equation is one where every term is dependent on the function or its derivatives. In mathematical terms, a second-order homogeneous linear differential equation can be expressed as: \( a f''(x) + b f'(x) + c f(x) = 0 \).
In our example, the equation \( f'' - f' - f = 0 \) fits perfectly into this form with coefficients \( a = 1 \), \( b = -1 \), and \( c = -1 \). Key characteristics of such equations are:
In our example, the equation \( f'' - f' - f = 0 \) fits perfectly into this form with coefficients \( a = 1 \), \( b = -1 \), and \( c = -1 \). Key characteristics of such equations are:
- The zero on the right side of the equation indicates that it is homogeneous. This sets them apart from non-homogeneous equations, which include an additional function on the right side.
- The solutions involve linear combinations of functions derived from solving the characteristic equation, which contains the coefficients of the different terms in the differential equation.
