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Magazine Chapter 6: Problem 1
Find the general solution. When the operator \(D\) is used, it is implied that the independent variable is \(x\). $$ \left(D^{2}-D-2\right) y=0. $$
Short Answer
Expert verified
The general solution is \( y(x) = C_1 e^{2x} + C_2 e^{-x} \).
Step by step solution
01
Write the Differential Equation with Operators
The given differential equation is \( (D^2 - D - 2)y = 0 \). Here, \( D \) is the differential operator, where \( D \equiv \frac{d}{dx} \). This equation can be rewritten as \( D^2 y - Dy - 2y = 0 \).
02
Find the Characteristic Equation
The characteristic equation is found by replacing \( D \) with \( r \) in the equation, leading to \( r^2 - r - 2 = 0 \). This is a quadratic equation.
03
Solve the Quadratic Equation
Solve the quadratic equation \( r^2 - r - 2 = 0 \) using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = -1 \), and \( c = -2 \).
04
Evaluate the Discriminant
The discriminant \( b^2 - 4ac \) is \( (-1)^2 - 4(1)(-2) = 1 + 8 = 9 \), which is positive, indicating two distinct real roots.
05
Find the Roots
Substitute into the quadratic formula: \( r = \frac{1 \pm \sqrt{9}}{2} \). This gives \( r = \frac{1 \pm 3}{2} \), resulting in the roots \( r_1 = 2 \) and \( r_2 = -1 \).
06
Write the General Solution
The general solution for a second-order differential equation with distinct roots \( r_1 \) and \( r_2 \) is \( y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \). Substitute \( r_1 = 2 \) and \( r_2 = -1 \) to get \( y(x) = C_1 e^{2x} + C_2 e^{-x} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Operator
A differential operator is a symbol that represents the process of differentiation. In the context of calculus, particularly in differential equations, the operator most commonly used is denoted by the letter \( D \). The differential operator \( D \) is defined as \( \frac{d}{dx} \), meaning it takes the derivative with respect to the variable \( x \). This simplifies the representation of differential equations by using algebraic notation.
- Example: \( Df \) means the derivative of the function \( f \) with respect to \( x \).
- \( D^2 \) would mean taking the derivative twice, or the second derivative.
- Applying these concepts, \( (D^2 - D - 2)y = 0 \) shows a second-order differential equation due to the highest power of \( D \).
Characteristic Equation
The characteristic equation is a crucial step when dealing with linear differential equations with constant coefficients. This equation arises by substituting the differential operator \( D \) with a variable, often \( r \). This essentially transforms the operator equation into an algebraic problem.
The specific characteristic equation for our differential equation \( (D^2 - D - 2)y = 0 \) is obtained by replacing \( D \) with \( r \), resulting in the equation \( r^2 - r - 2 = 0 \).
The specific characteristic equation for our differential equation \( (D^2 - D - 2)y = 0 \) is obtained by replacing \( D \) with \( r \), resulting in the equation \( r^2 - r - 2 = 0 \).
- This allows us to use algebraic techniques to solve for \( r \).
- The solution of this characteristic equation provides the roots that are pivotal in forming the general solution of the differential equation.
- Understanding how to transform and achieve the characteristic equation is key to solving these equations effectively.
Quadratic Formula
The quadratic formula is a mathematical tool used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). In our context, it helps find the roots of the characteristic equation we derived earlier. When you solve \( r^2 - r - 2 = 0 \), recognizing:
- \( a = 1 \)
- \( b = -1 \)
- \( c = -2 \)
- Calculate the discriminant, \( b^2 - 4ac = (-1)^2 - 4(1)(-2) = 9 \).
- The square root of 9 is 3, leading to two solutions: \( r = \frac{1 + 3}{2} = 2 \) and \( r = \frac{1 - 3}{2} = -1 \).
General Solution
The general solution of a differential equation provides a complete family of solutions. For second-order linear equations with constant coefficients, if the roots of the characteristic equation are distinct, the general solution can be expressed in a simple form.
From our solutions, we identified roots \( r_1 = 2 \) and \( r_2 = -1 \). The general solution is consequently given as:
From our solutions, we identified roots \( r_1 = 2 \) and \( r_2 = -1 \). The general solution is consequently given as:
- \( y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \)
- Substituting the values of the roots yields: \( y(x) = C_1 e^{2x} + C_2 e^{-x} \)
