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Chapter 10: Problem 21

Solve the initial-value problem. $$ y^{\prime}+2 y=1, y(0)=1 $$

Short Answer

Expert verified
The solution to the initial-value problem is \( y = \frac{1}{2}(1 + e^{-2x}) \).

Step by step solution

01

Identify the type of differential equation

The given differential equation is a first-order linear differential equation of the form: \( y' + 2y = 1 \).
02

Find the integrating factor

The standard form of a first-order linear differential equation is \( y' + P(x)y = Q(x) \). Here, \( P(x) = 2 \) and \( Q(x) = 1 \). The integrating factor \( \mu(x) \) is calculated as follows: \[ \mu(x) = e^{\int P(x) \, dx} = e^{2x} \].
03

Multiply the entire equation by the integrating factor

Multiplying the differential equation by the integrating factor \( e^{2x} \) gives: \[ e^{2x}y' + 2e^{2x}y = e^{2x} \].
04

Express the left-hand side as a derivative of a product

Notice that the left-hand side is the derivative of \( e^{2x}y \): \[ \frac{d}{dx}(e^{2x}y) = e^{2x} \].
05

Integrate both sides

Integrate both sides with respect to \( x \): \[ \int \frac{d}{dx}(e^{2x}y) \ dx = \int e^{2x} \ dx \]. This simplifies to: \[ e^{2x}y = \frac{1}{2} e^{2x} + C \].
06

Solve for y

Solve for \( y \) by dividing both sides by \( e^{2x} \): \[ y = \frac{1}{2} + Ce^{-2x} \].
07

Apply the initial condition

Use the initial condition \( y(0) = 1 \) to find \( C \): \[ 1 = \frac{1}{2} + C \Rightarrow C = \frac{1}{2} \].
08

Write the final solution

Substitute \( C \) back into the general solution: \[ y = \frac{1}{2} + \frac{1}{2} e^{-2x} \].

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Key Concepts

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

Integrating Factor
An integrating factor is a vital tool for solving first-order linear differential equations. It simplifies the equation to make it easier to solve. The form of a first-order linear differential equation is given by: \( y' + P(x) y = Q(x) \). Here, we identify \( P(x) \) and \( Q(x) \) from the provided equation. In this case, \( P(x) = 2 \) and \( Q(x) = 1 \).
To find the integrating factor \( \mu(x) \), use the formula:
  • \( \mu(x) = e^{\int P(x) \, dx} \).
For our problem, \( P(x) = 2 \). Therefore, the integrating factor becomes:
  • \( \mu(x) = e^{\int 2 \, dx} = e^{2x} \).
Multiply the entire differential equation by this integrating factor to transform it into a more easily solvable form.
Initial-Value Problem
An initial-value problem involves not only finding a general solution to a differential equation but also determining a particular solution that satisfies a given initial condition. In our example, the initial condition is \( y(0) = 1 \).
This initial condition means that when \( x = 0 \), \( y \) should equal 1. Applying this constraint helps us find the specific value of the constant involved in the general solution.
For this problem:
  • The general solution derived is \( y = \frac{1}{2} + Ce^{-2x} \).
  • To determine \( C \), we substitute \( x = 0 \) and \( y = 1 \) into the general solution:
    • \( 1 = \frac{1}{2} + C \).
    Solving for \( C \), we get \( C = \frac{1}{2} \).
This gives us the particular solution that satisfies the initial condition.
Differential Equation Solution
Solving a differential equation involves several steps, especially when dealing with first-order linear ones. Let's break down the process using our example:
  • First, recognize that the given equation \( y' + 2y = 1 \) is a first-order linear differential equation.
  • Next, find the integrating factor \( \mu(x) = e^{2x} \).
  • Multiplying the original equation with the integrating factor, we get: \( e^{2x} y' + 2e^{2x} y = e^{2x} \).
  • Notice that the left side is the derivative of a product: \( \frac{d}{dx}(e^{2x} y) \).
  • Integrate both sides: \( \int \frac{d}{dx}(e^{2x} y) \, dx = \int e^{2x} \, dx \).
  • Simplify to get: \( e^{2x} y = \frac{1}{2} e^{2x} + C \).
  • Solving for \( y \) gives: \( y = \frac{1}{2} + Ce^{-2x} \).
  • Finally, use the initial condition \( y(0) = 1 \) to find \( C \), resulting in the final particular solution:
      \( y = \frac{1}{2} + \frac{1}{2}e^{-2x} \).
This systematic method allows us to determine the solution to the differential equation efficiently.

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Most popular questions from this chapter

One or more initial conditions are given for each differential equation in the following exercises. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solutions. Include a \(y z\) -graph if one is not already provided. Always indicate the constant solutions on the \(t y\) -graph whether they are mentioned or not. $$ y^{\prime}=y-\frac{1}{4} y^{2}, y(0)=-1, y(0)=1 $$

Solve the given equation using an integrating factor. Take \(t>0\). $$ y^{\prime}+y=1 $$

In an experiment, a certain type of bacteria was being added to a culture at the rate of \(e^{.03 t}+2\) thousand bacteria per hour. Suppose that the bacteria grow at a rate proportional to the size of the culture at time \(t\), with constant of proportionality \(k=.45 .\) Let \(P(t)\) denote the number of bacteria in the culture at time \(t\). Find a differential equation satisfied by \(P(t)\).

According to the National Kidney Foundation, in 1997 more than 260,000 Americans suffered from chronic kidney failure and needed an artificial kidney (dialysis) to stay alive. (Source: The National Kidney Foundation, www.kidney.org.) When the kidneys fail, toxic waste products such as creatinine and urea build up in the blood. One way to remove these wastes is to use a process known as peritoneal dialysis, in which the patient's peritonium, or lining of the abdomen, is used as a filter. When the abdominal cavity is filled with a certain dialysate solution, the waste products in the blood filter through the peritonium into the solution. After a waiting period of several hours, the dialysate solution is drained out of the body along with the waste products. In one dialysis session, the abdomen of a patient with an elevated concentration of creatinine in the blood equal to 110 grams per liter was filled with two liters of a dialysate (containing no creatinine). Let \(f(t)\) denote the concentration of creatinine in the dialysate at time \(t\). The rate of change of \(f(t)\) is proportional to the difference between 110 (the maximum concentration that can be attained in the dialysate) and \(f(t)\). Thus, \(f(t)\) satisfies the differential equation $$ y^{\prime}=k(110-y) . $$ (a) Suppose that, at the end of a 4-hour dialysis session, the concentration in the dialysate was 75 grams per liter and it was rising at the rate of 10 grams per liter per hour. Find \(k\). (b) What is the rate of change of the concentration at the beginning of the dialysis session? By comparing with the rate at the end of the session, can you give a (simplistic) justification for draining and replacing the dialysate with a fresh solution after 4 hours of dialysis? [Hint: You do not need to solve the differential equation.]

A continuous annuity is a steady stream of money that is paid to some person. Such an annuity may be established, for example, by making an initial deposit in a savings account and then making steady withdrawals to pay the continuous annuity. Suppose that an initial deposit of $$\$ 5400$$ is made into a savings account that earns \(5 \frac{1}{2} \%\) interest compounded continuously, and immediately continuous withdrawals are begun at the rate of $$\$ 300$$ per year. Set up the differential equation that is satisfied by the amount \(f(t)\) of money in the account at time \(t .\) Sketch the solution.
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