Chapter 8: Problem 31
Give the first 5 terms of the series that is a solution to the given differential equation. $$y^{\prime}=3 y, \quad y(0)=1$$
Short Answer
Expert verified
The first 5 terms are: 1, 3x, \(\frac{9x^2}{2}\), \(\frac{9x^3}{2}\), \(\frac{27x^4}{8}\).
Step by step solution
01
Recognize the Problem Type
The differential equation given is a first-order linear ordinary differential equation of the form \( y' = ay \), with an initial condition \( y(0)=1 \).
02
Solve the Differential Equation
To solve \( y' = 3y \), we separate variables and integrate both sides. This gives us \( \frac{1}{y} dy = 3 dx \). Integrating both sides yields \( \ln|y| = 3x + C \). Exponentiating gives \( y = e^{3x+C} = Ce^{3x} \).
03
Apply Initial Condition
Use the initial condition to find \( C \). Substitute \( y(0) = 1 \) into \( y = Ce^{3x} \), which gives \( 1 = Ce^0 \) or \( C=1 \). Thus, the specific solution is \( y = e^{3x} \).
04
Find the Series Solution
The function \( e^{3x} \) can be expressed as a series using the exponential series expansion \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \). For \( e^{3x} \), replace \( x \) with \( 3x \): \( e^{3x} = \sum_{n=0}^{\infty} \frac{(3x)^n}{n!} \).
05
Compute the First 5 Terms
Calculate the first 5 terms of the series: \( \frac{(3x)^0}{0!} = 1 \), \( \frac{(3x)^1}{1!} = 3x \), \( \frac{(3x)^2}{2!} = \frac{9x^2}{2} \), \( \frac{(3x)^3}{3!} = \frac{27x^3}{6} = \frac{9x^3}{2} \), \( \frac{(3x)^4}{4!} = \frac{81x^4}{24} = \frac{27x^4}{8} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Linear Differential Equations
A first-order linear differential equation is a type of differential equation expressed in the form \( y' = ay + b \), where \( y' \) is the first derivative of \( y \) with respect to \( x \), and \( a \) and \( b \) are constants or functions of \( x \). In simpler terms, these equations involve the rate of change of a function and often appear in modeling real-world problems.
While it's possible to solve them using several methods, a common approach is separating variables. This allows us to integrate both sides to find a solution for \( y(x) \). In the given exercise, the equation is \( y' = 3y \), where \( a = 3 \), and \( b = 0 \).
By separating variables and integrating, we obtain a general solution which is typically in the form \( y = Ce^{ax} \), where \( C \) is a constant determined by initial conditions. The simplicity of first-order linear equations makes them a fundamental topic in understanding more complex systems in calculus.
While it's possible to solve them using several methods, a common approach is separating variables. This allows us to integrate both sides to find a solution for \( y(x) \). In the given exercise, the equation is \( y' = 3y \), where \( a = 3 \), and \( b = 0 \).
By separating variables and integrating, we obtain a general solution which is typically in the form \( y = Ce^{ax} \), where \( C \) is a constant determined by initial conditions. The simplicity of first-order linear equations makes them a fundamental topic in understanding more complex systems in calculus.
Series Solution
Finding a series solution involves expressing the solution of a differential equation as an infinite series. This method is particularly useful when the solution cannot be easily expressed in simple closed forms or in certain functions.
In this exercise, the differential equation \( y' = 3y \) has a solution \( y = e^{3x} \). To convert this function into a series, we use the Taylor series expansion for exponential functions, which is \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \).
By substituting \( 3x \) for \( x \) into this series, we get \( e^{3x} = \sum_{n=0}^{\infty} \frac{(3x)^n}{n!} \). This allows us to compute a polynomial approximation of the original function, using a number of terms as required. A series solution provides a way to approximate complex solutions and helps in understanding the behavior of the function over different intervals.
In this exercise, the differential equation \( y' = 3y \) has a solution \( y = e^{3x} \). To convert this function into a series, we use the Taylor series expansion for exponential functions, which is \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \).
By substituting \( 3x \) for \( x \) into this series, we get \( e^{3x} = \sum_{n=0}^{\infty} \frac{(3x)^n}{n!} \). This allows us to compute a polynomial approximation of the original function, using a number of terms as required. A series solution provides a way to approximate complex solutions and helps in understanding the behavior of the function over different intervals.
Exponential Function
The exponential function \( e^{x} \) is a critically important mathematical function characterized by a constant base, \( e \), raised to a variable exponent \( x \). It represents exponential growth or decay phenomena, common in natural and financial processes.
In the context of differential equations, exponential functions often appear as solutions to equations involving proportional change, like \( y' = 3y \). This particular equation, solved by separation and integration, gives the solution \( y = Ce^{3x} \), demonstrating exponential growth with a rate determined by the coefficient in the exponent.
The general form demonstrates how small changes consistently multiply the quantity; it’s a cornerstone concept in understanding dynamic systems in mathematics and science. The exponential expansion allows us to use series solutions to approximate and understand exponential behavior over intervals by using polynomial expressions.
In the context of differential equations, exponential functions often appear as solutions to equations involving proportional change, like \( y' = 3y \). This particular equation, solved by separation and integration, gives the solution \( y = Ce^{3x} \), demonstrating exponential growth with a rate determined by the coefficient in the exponent.
The general form demonstrates how small changes consistently multiply the quantity; it’s a cornerstone concept in understanding dynamic systems in mathematics and science. The exponential expansion allows us to use series solutions to approximate and understand exponential behavior over intervals by using polynomial expressions.
Initial Condition
An initial condition in differential equations provides necessary information to find a particular solution rather than merely a general one. It states the specific value of the function at a certain point, ensuring the solution satisfies conditions imposed by a real-world problem.
For the exercise \( y' = 3y \), the initial condition is \( y(0) = 1 \). Applying this condition allows us to find the precise constant \( C \) in the solution \( y = Ce^{3x} \). Determining \( C \) involves substituting the initial condition into the equation:
By applying the initial condition, we obtain \( y = e^{3x} \) as the particular solution. This illustrates how initial conditions are crucial in distinguishing specific solutions pertinent to scenarios modeled by differential equations.
For the exercise \( y' = 3y \), the initial condition is \( y(0) = 1 \). Applying this condition allows us to find the precise constant \( C \) in the solution \( y = Ce^{3x} \). Determining \( C \) involves substituting the initial condition into the equation:
- Substitute \( x = 0 \) and \( y(0) = 1 \) into \( y = Ce^{3x} \).
- Solving \( 1 = Ce^{0} \) gives \( C = 1 \).
By applying the initial condition, we obtain \( y = e^{3x} \) as the particular solution. This illustrates how initial conditions are crucial in distinguishing specific solutions pertinent to scenarios modeled by differential equations.
