/*! 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 6 . \(\frac{d y}{d x}=e^{-3 x}\)... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 6

. \(\frac{d y}{d x}=e^{-3 x}\)

Short Answer

Expert verified
The solution is \( y = -\frac{1}{3}e^{-3x} + C \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( \frac{dy}{dx} = e^{-3x} \), which is a first order, ordinary differential equation (ODE). Since it is expressed in terms of \( \frac{dy}{dx} \), it indicates that we need to find \( y \) by integrating the expression on the right-hand side.
02

Integrate the Differential Equation

To find \( y \), integrate both sides with respect to \( x \): \(\int \frac{d y}{d x} dx = \int e^{-3 x} dx.\)The left side integrates to \( y \), and the integral of \( e^{-3x} \) is calculated as: \(\int e^{-3 x} dx = -\frac{1}{3} e^{-3 x} + C,\) where \( C \) is the constant of integration.
03

Write Down the General Solution

The solution to the differential equation is:\(y = -\frac{1}{3} e^{-3x} + C.\)This represents the general solution, where \( C \) can be any real number, dependent on initial conditions if provided.

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Key Concepts

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

First Order Differential Equation
A first order differential equation is an equation that involves the derivatives of a function with respect to one variable. Here's what's important to note:
  • "First order" indicates that the equation involves the first derivative of the function, such as \( \frac{dy}{dx} \).
  • These equations describe the rate of change of a quantity, making them particularly useful for modeling various real-world processes like population growth, cooling of objects, or capacitor discharge in electrical circuits.
  • Solving these equations often involves finding an expression for the function in terms of its derivative.
In the given exercise, we have the first order differential equation \( \frac{dy}{dx} = e^{-3x} \). This tells us the rate at which \( y \) changes with respect to \( x \). Our task here is to determine \( y \) itself, by finding the expression that satisfies this rate of change.
Ordinary Differential Equation
An ordinary differential equation (ODE) is a differential equation containing one or more functions of one independent variable and its derivatives. Here are its key aspects:
  • "Ordinary" signifies that the equation involves functions of a single variable, unlike partial differential equations which involve partial derivatives with respect to multiple variables.
  • ODEs can be classified by their order and linearity; first order ODEs involve only the first derivative.
  • An ODE links the derivative of a function to the function itself, potentially including the initial or boundary conditions to find specific solutions.
In our example \( \frac{dy}{dx} = e^{-3x} \), we have an ODE since it connects the rate of change of \( y \) with the function \( y \) itself. By solving this equation, we aim to understand how \( y \) behaves as \( x \) changes, capturing its dependency on the specific form of the derivation.
Integration
Integration is a fundamental concept in calculus, crucial for solving differential equations. Here's how it fits into our problem:
  • Integration is the inverse process to differentiation. It allows us to find the original function given its derivative.
  • In the context of differential equations, integration is used to retrieve the dependent variable (in our case, \( y \)) from the expression involving its derivative.
  • When integrating, a constant of integration \( C \) is included to account for any initial conditions that might specify a particular solution.
To solve \( \frac{dy}{dx} = e^{-3x} \), we integrate \( e^{-3x} \) with respect to \( x \), resulting in \( y = -\frac{1}{3} e^{-3x} + C \). This integral process reverses the differentiation, providing the general solution of the ODE. Here, \( C \) can be adjusted based on particular initial conditions, thus tailoring the solution to fit specific scenarios.

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Most popular questions from this chapter

9\. A body moves along the \(x\) direction under the influence of the force \(F(t)=\cos 2 \pi t\), where, \(t\) is the time. (i) Write down the equation of motion. (ii) Find the solution that satisfies the initial conditions \(x(0)=0\) and \(\dot{x}(0)=1\)

\(\frac{d y}{d x}+\frac{3 y}{x}=3 x^{2} ; \quad y=\frac{x^{3}}{2}+\frac{c}{x^{3}}\)

Solve the initial value problems: 23\. \(\frac{d y}{d x}=\frac{x+y}{x} ; y(1)=2\)

Solve the initial value problems: 19\. \(\frac{d y}{d x}=\frac{y+2}{x-3} ; \quad y(0)=1\)

A third-order process \(\mathrm{A}+2 \mathrm{~B} \rightarrow\) products has rate equation $$ \frac{d x}{d t}=k(a-x)(b-2 x)^{2} $$ where \(a\) and \(b\) are the initial amounts of \(\mathrm{A}\) and \(\mathrm{B}\), respectively. Show that the solution that satisfies the initial condition \(x(0)=0\) is given by $$ k t=\frac{1}{(2 a-b)^{2}} \ln \frac{a(b-2 x)}{b(a-x)}+\frac{2 x}{b(2 a-b)(b-2 x)} $$
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