Chapter 6: Problem 2
Find the general solution. When the operator \(D\) is used, it is implied that the independent variable is \(x\). $$ \left(D^{2}+3 D\right) y=0. $$
Short Answer
Expert verified
The general solution is \( y = C_1 + C_2e^{-3x} \).
Step by step solution
01
Identify the Differential Equation
The problem provides a differential equation given as \( \left(D^{2}+3 D\right)y = 0 \). Here, \( D \) is the differential operator, meaning it represents differentiation with respect to the independent variable, \( x \).
02
Apply the Characteristic Equation Approach
Rewrite the differential equation in terms of the operator \( D \), treating it as an algebraic expression: \( (D^2 + 3D)y = 0 \). The characteristic equation associated with this operator is obtained by substituting \( y = e^{rx} \) into the differential equation, which leads to \( (r^2 + 3r) = 0 \).
03
Solve the Characteristic Equation
The characteristic equation \( r^2 + 3r = 0 \) can be solved by factoring. Factoring the equation gives \( r(r + 3) = 0 \). This implies \( r = 0 \) or \( r = -3 \).
04
Write the General Solution
The general solution of a differential equation with characteristic roots \( r_1 \) and \( r_2 \) is given by \( y = C_1e^{r_1x} + C_2e^{r_2x} \). Substituting the roots \( r_1 = 0 \) and \( r_2 = -3 \) into the formula, we obtain \( y = C_1 + C_2e^{-3x} \), where \( C_1 \) and \( C_2 \) are arbitrary constants.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Operators
Differential operators are a fundamental concept when dealing with differential equations. In mathematics, a differential operator refers to an operator defined as a function of the differentiation operator. The most common differential operator is the derivative, denoted by \( D \) or \( \frac{d}{dx} \), where the operation is performed with respect to a variable, often \( x \). This operator is used to denote taking the derivative of a function.
For example, if \( Df(x) = \frac{df}{dx} \), then applying \( D \) once to a function involves differentiating it once. Higher-order differential operators like \( D^2 \) involve taking the derivative multiple times. In our exercise, \( D^2 + 3D \) operates on \( y \), implying two successes of differentiation and a combination of operations.
The operator approach simplifies handling differential equations by turning them into algebraic equations, making them easier to solve.
For example, if \( Df(x) = \frac{df}{dx} \), then applying \( D \) once to a function involves differentiating it once. Higher-order differential operators like \( D^2 \) involve taking the derivative multiple times. In our exercise, \( D^2 + 3D \) operates on \( y \), implying two successes of differentiation and a combination of operations.
The operator approach simplifies handling differential equations by turning them into algebraic equations, making them easier to solve.
Characteristic Equation
The characteristic equation is a powerful tool used to solve linear differential equations. It simplifies the process of finding the solution by transforming a differential equation into an algebraic equation. This is possible by assuming the solution has an exponential form.
For instance, substituting \( y = e^{rx} \) into the differential equation allows us to substitute differentiation operations as algebraic multiplication involving \( r \). So, the differential equation \( (D^2 + 3D)y = 0 \) becomes the characteristic equation \( r^2 + 3r = 0 \).
The characteristic equation reduces the need to directly solve complex differential equations by finding the roots \( r \) of the polynomial, turning the differential problem into an algebraic one.
For instance, substituting \( y = e^{rx} \) into the differential equation allows us to substitute differentiation operations as algebraic multiplication involving \( r \). So, the differential equation \( (D^2 + 3D)y = 0 \) becomes the characteristic equation \( r^2 + 3r = 0 \).
The characteristic equation reduces the need to directly solve complex differential equations by finding the roots \( r \) of the polynomial, turning the differential problem into an algebraic one.
General Solution
The general solution of a differential equation provides a formula that represents all possible solutions. For linear equations with constant coefficients, the solution often involves exponential functions, especially when characteristic roots are real numbers.
After finding the roots \( r_1 \) and \( r_2 \) from the characteristic equation, the general solution is typically expressed as \( y = C_1e^{r_1x} + C_2e^{r_2x} \), where \( C_1 \) and \( C_2 \) are arbitrary constants determined by boundary or initial conditions. In simpler terms, for our example with roots \( r_1 = 0 \) and \( r_2 = -3 \), the general solution becomes:
After finding the roots \( r_1 \) and \( r_2 \) from the characteristic equation, the general solution is typically expressed as \( y = C_1e^{r_1x} + C_2e^{r_2x} \), where \( C_1 \) and \( C_2 \) are arbitrary constants determined by boundary or initial conditions. In simpler terms, for our example with roots \( r_1 = 0 \) and \( r_2 = -3 \), the general solution becomes:
- \( y = C_1 + C_2e^{-3x} \)
Factoring Quadratic Equations
Factoring quadratic equations is a crucial step in solving many algebraic and differential problems. In the context of finding the characteristic roots, factoring is used to simplify a quadratic into terms whose product is the quadratic equation itself.
For the characteristic equation \( r^2 + 3r = 0 \), the process involves identifying common factors and applying the zero-product property:
For the characteristic equation \( r^2 + 3r = 0 \), the process involves identifying common factors and applying the zero-product property:
- Factoring gives \( r(r + 3) = 0 \)
- This yields the solutions \( r = 0 \) or \( r = -3 \)