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Chapter 10: Problem 4

Show that the function \(f(t)=\left(e^{-t}+1\right)^{-1}\) satisfies \(y^{\prime}+y^{2}=y, y(0)=\frac{1}{2}\).

Short Answer

Expert verified
Function \( f(t) = \frac{1}{e^{-t} + 1} \) satisfies the differential equation and initial condition.

Step by step solution

01

- Find the first derivative of the function

To show that the function satisfies the differential equation, first find the derivative of the function. Given function is: \[ f(t) = \frac{1}{e^{-t} + 1} \]We need to use the chain rule to differentiate this. Letting \[ u = e^{-t} + 1 \], we get the derivative as: \[ f'(t) = - \frac{d}{dt} \frac{1}{u} = - \frac{1 \times -e^{-t}}{(e^{-t}+1)^2} = \frac{e^{-t}}{(e^{-t} + 1)^2} \].
02

- Substitute function into the differential equation

Given the differential equation: \[ y' + y^2 = y \]. Substitute \[ y = f(t) = \frac{1}{e^{-t} + 1} \] and \[ y' = \frac{e^{-t}}{(e^{-t} + 1)^2} \]: \[ \frac{e^{-t}}{(e^{-t} + 1)^2} + \frac{1}{(e^{-t} + 1)^2} = \frac{1}{e^{-t} + 1} \].
03

- Simplify the equation

Combine the fractions on the left side of the differential equation: \[ \frac{e^{-t}}{(e^{-t} + 1)^2} + \frac{1}{(e^{-t} + 1)^2} = \frac{e^{-t} + 1}{(e^{-t} + 1)^2} = \frac{1}{e^{-t} + 1} \]. The left side simplifies to match the right side.
04

- Verify the initial condition

Check the initial condition \( y(0) = \frac{1}{2} \): When \( t = 0 \), \[ f(0) = \frac{1}{e^{0} + 1} = \frac{1}{2} \]. This confirms that \( y(0) = \frac{1}{2} \).

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Key Concepts

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

chain rule
The chain rule is a crucial concept in calculus for differentiating composite functions. It allows us to find the derivative of a function by breaking it down into simpler parts.
The chain rule states that if you have a composite function, such as \( h(x) = f(g(x)) \), the derivative is given by: \[ h'(x) = f'(g(x)) \times g'(x) \]In our exercise, we used the chain rule to differentiate \( f(t) = \frac{1}{e^{-t} + 1} \).
Here's how we applied it:
  • First, we let \( u = e^{-t} + 1 \). This makes \( f(t) \) easier to handle.
  • Next, we differentiate \( f(t) \) with respect to \( u \) and then multiply by the derivative of \( u \) with respect to \( t \).
  • This simplifies to \( f'(t) = \frac{e^{-t}}{(e^{-t} + 1)^2} \).
The chain rule unravels complex functions step-by-step, making differentiation manageable and systematic.
initial conditions
Initial conditions specify the value of the function at a particular point and are essential for solving differential equations.
They provide a specific solution from the general solution of the differential equation. For instance, the exercise states the initial condition as \( y(0) = \frac{1}{2} \).
This tells us that when \( t = 0 \), the function \( y(t) \) must equal \( \frac{1}{2} \).
  • Initial conditions help us determine constants of integration.
  • They ensure the solution is uniquely defined.

In our exercise, we confirmed the initial condition by substituting \( t = 0 \) into the function \( f(t) \) and obtaining \[ f(0) = \frac{1}{e^{0} + 1} = \frac{1}{2} \]Initial conditions are crucial for the accuracy and applicability of the solutions in real-world scenarios.
first derivative
The first derivative represents the rate of change of a function concerning its variable. It's a fundamental concept in understanding the behavior of functions.
In our exercise, we computed the first derivative of \( f(t) \) to check if it satisfies the given differential equation. The function is \( f(t) = \frac{1}{e^{-t} + 1} \).
Using the chain rule, we found the first derivative:
\[ f'(t) = \frac{e^{-t}}{(e^{-t} + 1)^2} \].
  • The first derivative tells us how \( f(t) \) changes as \( t \) changes.
  • It plays a key role in differential equations, providing the necessary information to check if a function is a solution.

Understanding the first derivative helps us analyze the function's behavior, including its growth and decay rates.
function substitution
Function substitution is a technique used to simplify differential equations by replacing a function with another expression.
It helps in making complex equations more manageable and easier to solve.
In our exercise, we substituted \( y \) and \( y' \) into the differential equation to verify the solution.
  • Given \( y = \frac{1}{e^{-t} + 1} \) and \( y' = \frac{e^{-t}}{(e^{-t} + 1)^2} \), we substituted them back into \ y' + y^2 = y \.
  • Combining fractions on the left side simplifies the equation: \[ \frac{e^{-t}}{(e^{-t} + 1)^2} + \frac{1}{(e^{-t} + 1)^2} = \frac{e^{-t} + 1}{(e^{-t} + 1)^2} = \frac{1}{e^{-t} + 1} \]
Function substitution helped simplify and verify that the function satisfies the differential equation.
It's a powerful tool in solving and manipulating differential equations effectively.

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Most popular questions from this chapter

Let \(q=f(p)\) be the demand function for a certain commodity, where \(q\) is the demand quantity and \(p\) the price of 1 unit. In Section 5.3, we defined the elasticity of demand as $$ E(p)=\frac{-p f^{\prime}(p)}{f(p)} $$ (a) Find a differential equation satisfied by the demand function if the elasticity of demand is a linear function of price given by \(E(p)=p+1\). (b) Find the demand function in part (a), given \(f(1)=100\).

The Federal Housing Finance Board reported that the national average price of a new one-family house in 2012 was $$\$ 278,900$$. At the same time, the average interest rate on a conventional 30 -year fixedrate mortgage was \(3.1 \% .\) A person purchased a home at the average price, paid a down payment equal to \(10 \%\) of the purchase price, and financed the remaining balance with a 30 -year fixed-rate mortgage. Assume that the person makes payments continuously at a constant annual rate \(A\) and that interest is compounded continuously at the rate of \(3.1 \% .\) (Source: The Federal Housing Finance Board, www.fhfb.gov.) (a) Set up a differential equation that is satisfied by the amount \(f(t)\) of money owed on the mortgage at time \(t\) (b) Determine \(A\), the rate of annual payments, that is required to pay off the loan in 30 years. What will the monthly payments be? (c) Determine the total interest paid during the 30 -year term mortgage.

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