/*! 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 29 Consider the initial-value probl... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 29

Consider the initial-value problem $$y^{\prime}=-\frac{y}{1+t}+10, \quad y(0)=50.$$ (a) Is the solution increasing or decreasing when \(t=0 ?\) [Hint: Compute \(\left.y^{\prime}(0) .\right]\) (b) Find the solution and plot it for \(0 \leq t \leq 4.\)

Short Answer

Expert verified
The solution is decreasing at \(t=0\) with \(y^{\prime}(0) = -40\); the solution is \(y(t) = 5(t+2) + \frac{45}{1+t}\).

Step by step solution

01

Compute the initial derivative

To determine if the solution is increasing or decreasing at \(t=0\), calculate \(y^{\prime}(0)\). Substitute \(t = 0\) and \(y(0) = 50\) into the given differential equation: \[ y^{\prime} = -\frac{y}{1+t} +10 \ y^{\prime}(0) = -\frac{50}{1+0} + 10 = -50 + 10 = -40 \] Since \(y^{\prime}(0) = -40\), the solution is decreasing when \(t=0\).
02

Solve the differential equation

To find the solution, we use the method of integrating factors. Rewrite the equation as \[ y^{\prime} + \frac{y}{1+t} = 10. \] Identify the integrating factor, \( \mu(t) = e^{\int \frac{1}{1+t} dt} = e^{\ln|1+t|} = |1+t| \). For \(t \geq 0\), \( |1+t| = 1+t \). Multiply through by the integrating factor: \[ (1+t) y^{\prime} + y = 10(1+t). \] The left side is the derivative of \((1+t)y\), so integrate both sides: \[ \int d[(1+t)y] = \int 10(1+t) \, dt \ (1+t)y = 10 \int (1+t) \, dt = 10 \left(\frac{1+t^2}{2} + t \right) + C \ (1+t)y = 10\left(\frac{t^2}{2} + t + \frac{1}{2} \right) + C \ y(t) = \frac{10}{1+t} \left( \frac{t^2}{2} + t + \frac{1}{2} \right) + \frac{C}{1+t}. \]
03

Apply initial condition

Apply the initial condition \(y(0)=50\) to find \(C\). Substitute \(t = 0\) and \(y(0) = 50\) into the solution: \[ 50 = 10\left(\frac{0^2}{2} + 0 + \frac{1}{2}\right) + C \ 50 = 10 \cdot 0.5 + C \ 50 = 5 + C \ C = 45. \] Therefore, the solution is \[ y(t) = \frac{10(t^2 + 2t + 1) + 90}{2(1+t)} = \frac{10(t^2 + 2t + 1) + 90}{2(1+t)}. \]
04

Simplify the solution

Simplify the expression for the general solution: \[ y(t) = 5(t+2) + \frac{45}{1+t}. \]
05

Plot the solution

Plot the solution \(y(t) = 5(t+2) + \frac{45}{1+t}\) for \(0 \leq t \leq 4\) using a graphing tool or software to visualize the behavior of the solution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Differential Equations
Differential equations are mathematical equations that relate a function with its derivatives. These equations describe how a certain quantity changes over time. In this exercise, we are dealing with a first-order differential equation, which involves the first derivative of the unknown function, noted as \( y' \). The equation we are solving looks like this: \( y^{\text{prime}}=-\frac{y}{1+t}+10 \). This equation tells us how \( y \) changes with respect to \( t \).
Integrating Factors
An integrating factor is a function used to simplify solving a linear differential equation. By multiplying both sides of the equation by this factor, we can transform it into a simpler form that can be integrated directly. In our exercise, the integrating factor is \( \mu(t) = e^{\int \frac{1}{1+t} \, dt} = e^{\ln|1+t|} = |1+t| \). For non-negative \( t \), this simplifies to \( 1+t \). Multiplying the entire equation by \( 1+t \), we can then integrate both sides easily.
Calculus Solutions
Solving differential equations often involves using calculus techniques such as integration. Once we have multiplied by the integrating factor and simplified the equation, we use integration to find the solution. In the initial-value problem given, we integrate both sides to get: \( \int d[(1+t)y] = \int 10(1+t) \, dt \). Then, substituting the limits of integration based on initial conditions, we find the particular solution.
Initial Conditions
Initial conditions are values given for the function and its derivatives at a specific point. They allow us to find particular solutions to differential equations. Here, our initial condition is \( y(0) = 50 \). This means that when \( t = 0 \), \( y \) should equal 50. By substituting this into our general solution formula, we can solve for the constant \( C \). So, we substitute \( t = 0 \) and \( y(0) = 50 \), and find that \( C \) matches the initial condition, which gives us the unique solution to our problem.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find an integrating factor for each equation. Take \(t>0\). $$y^{\prime}=t^{2}(y+1)$$

Radium 226 is a radioactive substance with a decay constant .00043. Suppose that radium 226 is being continuously added to an initially empty container at a constant rate of 3 milligrams per year. Let \(P(t)\) denote the number of grams of radium 226 remaining in the container after years (a) Find an initial-value problem satisfied by \(P(t).\) (b) Solve the initial-value problem for \(P(t).\) (c) What is the limit of the amount of radium 226 in the container as \(t\) tends to infinity?

Solve the initial-value problem. $$y^{\prime}=2(10-y), y(0)=1$$

Review concepts that are important in this section. In each exercise, sketch the graph of a function with the stated properties. Domain: \(0 \leq t \leq 7 ;(0,2)\) is on the graph; the slope is always positive, the slope becomes more positive as \(t\) increases from 0 to 3, and the slope becomes less positive as \(t\) increases from 3 to 7.

The air in a crowded room full of people contains \(.25 \%\) carbon dioxide \(\left(\mathrm{CO}_{2}\right) .\) An air conditioner is turned on that blows fresh air into the room at the rate of 500 cubic feet per minute. The fresh air mixes with the stale air, and the mixture leaves the room at the rate of 500 cubic feet per minute. The fresh air contains \(.01 \% \mathrm{CO}_{2},\) and the room has a volume of 2500 cubic feet. (a) Find a differential equation satisfied by the amount \(f(t)\) of \(\mathrm{CO}_{2}\) in the room at time \(t.\) (b) The model developed in part (a) ignores the \(\mathrm{CO}_{2}\) produced by the respiration of the people in the room. Suppose that the people generate .08 cubic foot of \(\mathrm{CO}_{2}\) per minute. Modify the differential equation in part (a) to take into account this additional source of \(\mathrm{CO}_{2}.\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.