/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Solve the following initial valu... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 22

Solve the following initial value problem for \(u\) as a function of \(t\): \(\frac{d u}{d t}+\frac{k}{m} u=0 \quad(k \text { and } m \text { positive constants }), \quad u(0)=u_{0}\) a. as a first-order linear equation. b. as a separable equation.

Short Answer

Expert verified
The solution is \( u(t) = u_0 e^{-\frac{k}{m} t} \) using both methods (linear and separable).

Step by step solution

01

Identify the Form of the Equation

The given differential equation \( \frac{du}{dt} + \frac{k}{m} u = 0 \) is a first-order linear differential equation. The standard form of a first-order linear equation is \( \frac{du}{dt} + P(t)u = Q(t) \). Here, \( P(t) = \frac{k}{m} \) and \( Q(t) = 0 \). It is also separable by nature, which means it can be solved using separation of variables.
02

Solution as a First-Order Linear Equation

For a first-order linear equation \( \frac{du}{dt} + P(t) u = Q(t) \), we use an integrating factor \( \mu(t) = e^{\int P(t) \, dt} \). In this case, \( P(t) = \frac{k}{m} \), so \( \mu(t) = e^{\int \frac{k}{m} \, dt} = e^{\frac{k}{m}t} \). Multiply the entire differential equation by \( \mu(t) \) to get \( e^{\frac{k}{m}t} \frac{du}{dt} + e^{\frac{k}{m}t} \frac{k}{m} u = 0 \). Integrating both sides with respect to \( t \) gives \( e^{\frac{k}{m}t} u = C \), and thus \( u = Ce^{-\frac{k}{m}t} \). Applying the initial condition \( u(0) = u_0 \), we get \( C = u_0 \). Therefore, \( u(t) = u_0 e^{-\frac{k}{m}t} \).
03

Solution as a Separable Equation

The equation \( \frac{du}{dt} + \frac{k}{m} u = 0 \) can be rewritten as \( \frac{du}{dt} = -\frac{k}{m} u \). This form allows separation of variables: \( \frac{du}{u} = -\frac{k}{m} dt \). Integrating both sides, we have \( \ln|u| = -\frac{k}{m} t + C \). By exponentiation, \( u = e^{C} e^{-\frac{k}{m} t} \). Let \( e^{C} = C_1 \), then \( u = C_1 e^{-\frac{k}{m} t} \). Applying the initial condition \( u(0) = u_0 \), we find \( C_1 = u_0 \), yielding \( u(t) = u_0 e^{-\frac{k}{m} t} \).
04

Verify and Interpret the Solution

Both methods lead to the same solution \( u(t) = u_0 e^{-\frac{k}{m} t} \). This function describes an exponential decay over time, depending on the initial value \( u_0 \), the decay constant \( k \), and the mass \( m \). Convergence to zero occurs as time increases.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First-Order Linear Differential Equation
A first-order linear differential equation is an equation that involves a function and its first derivative.
These equations have the general form \( \frac{du}{dt} + P(t)u = Q(t) \). Here, \( P(t) \) and \( Q(t) \) can be any functions of \( t \), but they are often constants, especially in simpler problems.
  • **Structure**: The structure of these equations allows them to be solved using systematic methods.
  • **Integrating Factor**: An important method for solving these equations is using an integrating factor. This is a function, often exponential, that transforms the equation into an easily integrable form.
  • **Initial Value Problems**: These equations often come with an initial condition, like \( u(0) = u_0 \), which allows us to find the constant of integration and thus a unique solution.
Steps to solve include identifying the parts of the equation, finding an integrating factor \( \mu(t) = e^{\int P(t) dt} \), and using it to solve for \( u(t) \). In the given problem, the integrating factor is \( e^{\frac{k}{m}t} \), which on multiplication and integrating, gives rise to the solution.
Separation of Variables
Separation of variables is another powerful technique used to solve differential equations, especially when the equation is already in or can be rearranged into a separable form.
  • **Separable Form**: A differential equation is separable if it can be expressed as \( \frac{du}{dt} = g(t) u = f(t) u \), allowing the variables to be separated into \( \int \frac{1}{u} du = \int -\frac{k}{m} dt \).
  • **Integration**: Once the variables are separated, the next step is to integrate both sides. This will typically involve integrating functions like \( \ln(u) \) for the \( u \) side, and linear terms for the \( t \) side.
  • **Exponentiation**: After integration, you would often exponentiate both sides to solve for \( u \), especially when the equation originally involved logarithmic integration.
  • **Initial Conditions**: Use these to determine any constants of integration. For the given problem, the condition \( u(0) = u_0 \) is applied after integration.
In the exercise, separation yields the same solution \( u(t) = u_0 e^{-\frac{k}{m}t} \), confirming the validity of both methods.
Exponential Decay
In mathematics, exponential decay describes the process of reducing an amount by a consistent percentage rate over a period of time.
  • **Decay Formula**: The general form is \( u(t) = u_0 e^{-kt} \), but in physical problems, you often have additional parameters, like mass \( m \) in this problem.
  • **Decay Constant**: Here, \( k/m \) acts as the decay constant. It dictates the rate of decay; larger values mean a quicker decrease.
  • **Time Dependency**: As \( t \) increases, \( e^{-\frac{k}{m}t} \) approaches zero, indicating how the initial amount decays away over time.
  • **Applications**: Exponential decay models various real-world processes like radioactive decay, cooling of a hot object, or depreciation of value over time.
For the given problem, the solution \( u(t) = u_0 e^{-\frac{k}{m}t} \) elegantly captures the idea of exponential decay, where the quantity \( u \) decreases as time progresses, depending heavily on the constants \( k \) and \( m \).

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