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Chapter 9: Problem 18

Find the general solution of each differential equation or state that the differential equation is not separable. If the exercise says "and check," verify that your answer is a solution. $$ y^{\prime}=6 e^{-2 x} $$

Short Answer

Expert verified
The general solution is \( y = -3e^{-2x} + C \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( y' = 6e^{-2x} \). This is a first-order ordinary differential equation and appears to be separable, i.e., it can be expressed in the form \( \frac{dy}{dx} = g(x) \).
02

Rewrite as an Integral

To find the general solution, we integrate both sides with respect to \( x \). The equation becomes \( \int \, dy = \int 6e^{-2x} \, dx \).
03

Solve the Integral for the Right-Hand Side

Let's integrate the right side: \( \int 6e^{-2x} \, dx \).Use the substitution: let \( u = -2x \), then \( du = -2dx \) or \( dx = -\frac{1}{2}du \).The integral becomes \( 6 \int e^{u} \left(-\frac{1}{2}\right) du = -3 \int e^{u} \, du \).The integral of \( e^u \) is itself, hence, \( -3e^u + C = -3e^{-2x} + C \) where \( C \) is the integration constant.
04

Solve for y(x)

Integrating the left side gives \( y(x) = \int dy \), which results in \( y(x) = y = -3e^{-2x} + C \). Thus, the general solution to the differential equation is \( y(x) = -3e^{-2x} + C \), where \( C \) is an arbitrary constant.

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Key Concepts

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

General Solution
A general solution to a differential equation includes all possible solutions that satisfy the equation.
This is different from a particular solution, which is specific and determined by initial conditions.
The arbitrary constant, usually denoted as \( C \), represents the integration constant that emerges from solving differential equations.
In our case, the general solution to the equation \( y' = 6e^{-2x} \) is \( y(x) = -3e^{-2x} + C \).
This expression showcases the family of curves that solve the differential equation, where each value of \( C \) corresponds to a different curve.
To obtain a specific function \( y(x) \), an initial condition is necessary.
First-Order Ordinary Differential Equation
First-order ordinary differential equations involve the first derivative of the unknown function, but not higher derivatives.
These equations typically take the form \( \frac{dy}{dx} = f(x, y) \).
In our specific example, the equation \( y' = 6e^{-2x} \) is a first-order ordinary differential equation because it only includes \( y' \), the first derivative of \( y \).
Such equations are prevalent in modeling real-world phenomena, including exponential growth and decay, logistic processes, and motion under constant forces.
This type of differential equation is prominent because it is often easier to solve compared to higher-order differential equations.
Separable Differential Equation
Separable differential equations are those that can be rewritten to separate the variables on different sides of the equation.
They generally have the form \( g(y) \, dy = f(x) \, dx \) and are solvable by integration.
Our example, \( y' = 6e^{-2x} \), is separable because \( \frac{dy}{dx} = 6e^{-2x} \) can be rewritten as \( dy = 6e^{-2x} \, dx \).
This separation allows us to integrate both sides independently, simplifying the solving process.
By doing this, we transform the problem from a differential equation to finding the antiderivative, which is much easier to handle.

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Most popular questions from this chapter

Solve each differential equation and initial condition and verify that your answer satisfies both the differential equation and the initial condition. $$\left\\{\begin{array}{l} y^{\prime}=a x y \\ y(0)=4 \end{array} \quad(\text { for constant } a>0)\right.$$

A medical examiner called to the scene of a murder will usually take the temperature of the body. A corpse cools at a rate proportional to the difference between its temperature and the temperature of the room. If \(y(t)\) is the temperature (in degrees Fahrenheit) of the body \(t\) hours after the murder, and if the room temperature is \(70^{\circ},\) then \(y\) satisfies $$ \begin{aligned} y^{\prime} &=-0.32(y-70) \\ y(0) &=98.6 \text { (body temperature initially } \left.98.6^{\circ}\right) \end{aligned}$$ a. Solve this differential equation and initial condition. b. Use your answer to part (a) to estimate how long ago the murder took place if the temperature of the \text { body when it was discovered was } 80^{\circ} .

An annuity is a fund into which one makes equal payments at regular intervals. If the fund earns interest at rate \(r\) compounded continuously, and deposits are made continuously at the rate of \(d\) dollars per year (a continuous annuity), then the value \(y(t)\) of the fund after \(t\) years satisfies the differential equation \(y^{\prime}=d+r y .\) (Do you see why?) Solve the differential equation in the preceding instructions for the continuous annuity \(y(t)\) with deposit rate \(d=\$ 1000\) and continuous interest rate \(r=0.05,\) subject to the initial condition \(y(0)=0\) (zero initial value).

For each initial value problem, use an Euler's method graphing calculator program to find the approximate solution at the stated \(x\) -value, using 50 segments. [Hint: Use an interval that begins at the initial \(x\) -value and ends at the stated \(x\) -value. \(\frac{d y}{d x}=(x-y)^{2}\) \(y(2)=0\) Approximate the solution at \(x=2.8\)

For each initial value problem: a. Use an Euler's method graphing calculator program to find the estimate for \(y(2)\). Use the interval [0,2] with \(n=50\) segments. b. Solve the differential equation and initial condition exactly by separating variables or using an integrating factor. c. Evaluate the solution that you found in part (b) at \(x=2\). Compare this actual value of \(y(2)\) with the estimate of \(y(2)\) that you found in part (a). $$ \begin{array}{l} y^{\prime}=\frac{x}{y} \\ y(0)=1 \end{array} $$
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