/*! 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 27 Solve the initial-value problem.... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 27

Solve the initial-value problem. \(\frac{d u}{d t}=\sqrt{t}+\frac{2}{\sqrt{t}}, \quad t>0, u(1)=5\)

Short Answer

Expert verified
The solution is \( u(t) = \frac{2}{3}t^{\frac{3}{2}} + 4t^{\frac{1}{2}} - \frac{7}{3} \).

Step by step solution

01

Integrate the Differential Equation

To solve the given differential equation \( \frac{du}{dt} = \sqrt{t} + \frac{2}{\sqrt{t}} \), we integrate both sides with respect to \( t \). The integral of \( \sqrt{t} \) is \( \frac{2}{3}t^{\frac{3}{2}} \), and the integral of \( \frac{2}{\sqrt{t}} \) is \( 4t^{\frac{1}{2}} \). Therefore, \( u(t) = \frac{2}{3}t^{\frac{3}{2}} + 4t^{\frac{1}{2}} + C \), where \( C \) is the constant of integration.
02

Apply Initial Condition

Given the initial condition \( u(1) = 5 \), substitute \( t = 1 \) and \( u = 5 \) into the equation \( u(t) = \frac{2}{3}t^{\frac{3}{2}} + 4t^{\frac{1}{2}} + C \). This gives us \( 5 = \frac{2}{3}(1)^{\frac{3}{2}} + 4(1)^{\frac{1}{2}} + C \). Simplifying, we find \( 5 = \frac{2}{3} + 4 + C \), which leads to \( C = \frac{5}{3} - 4 \). Therefore, \( C = \frac{5}{3} - \frac{12}{3} = -\frac{7}{3} \).
03

Write the Solution

Substitute \( C = -\frac{7}{3} \) back into the general solution \( u(t) = \frac{2}{3}t^{\frac{3}{2}} + 4t^{\frac{1}{2}} + C \). Thus, the particular solution satisfying both the differential equation and initial condition is \( u(t) = \frac{2}{3}t^{\frac{3}{2}} + 4t^{\frac{1}{2}} - \frac{7}{3} \).

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.

Understanding Differential Equations
A differential equation is a mathematical equation that relates a function with its derivatives. In simpler terms, it tells us how a function changes over time or another variable. In our original exercise, the differential equation is given as \( \frac{du}{dt} = \sqrt{t} + \frac{2}{\sqrt{t}} \). This equation explains how the function \( u \) changes with respect to time \( t \).
This equation is of the first order since it involves the first derivative \( \frac{du}{dt} \). To solve this, we need to express \( u \) purely in terms of \( t \), which means finding the antiderivative or integrating the given expression with respect to \( t \).
Integration Explained
Integration is the process of finding the integral, or antiderivative, of a function. It is the reverse process of differentiation. In our exercise, the goal is to find \( u(t) \) by integrating the function \( \sqrt{t} + \frac{2}{\sqrt{t}} \).
  • The integral of \( \sqrt{t} \) or \( t^{1/2} \) is \( \frac{2}{3}t^{3/2} \). This follows the power rule for integration.
  • Similarly, the integral of \( \frac{2}{\sqrt{t}} \) or \( 2t^{-1/2} \) is \( 4t^{1/2} \).
When we integrate the entire right side, we get \( u(t) = \frac{2}{3}t^{3/2} + 4t^{1/2} + C \), where \( C \) is the constant of integration that we'll discuss next.
The Constant of Integration
When integrating, we must always add a constant of integration, denoted by \( C \), to the solution. This constant is crucial because it accounts for the fact that differentiating a constant gives zero, thus it can be any value. In solving differential equations, the specific value of \( C \) is often unknown unless additional information, like an initial condition, is provided.
In our solution:
  • The expression \( u(t) = \frac{2}{3}t^{3/2} + 4t^{1/2} + C \) represents a family of functions that differ only by the value of \( C \).
  • The initial condition \( u(1) = 5 \) helps to determine this value of \( C \).
Initial Condition Defined
An initial condition is a value that specifies the state of a function at a particular point. It is often given in problems involving differential equations to find a particular solution rather than a general one. In our problem, the initial condition is \( u(1) = 5 \). This means that when \( t = 1 \), the value of \( u \) is \( 5 \).
By substituting these values into our integrated function \( u(t) = \frac{2}{3}t^{3/2} + 4t^{1/2} + C \), we find:
  • \( 5 = \frac{2}{3}(1)^{3/2} + 4(1)^{1/2} + C \).
  • Simplifying this equation, we calculate \( C \) as \( -\frac{7}{3} \).
Finally, substituting \( C \) back into our function gives the particular solution: \( u(t) = \frac{2}{3}t^{3/2} + 4t^{1/2} - \frac{7}{3} \). This solution satisfies both the differential equation and the initial condition provided.

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

A rectangular storage container with an open top is to have a volume of 10 \(\mathrm{m}^{3} .\) The length of its base is twice the width. Material for the base costs \(\$ 10\) per square meter. Material for the sides costs \(\$ 6\) per square meter. Find the cost of materials for the cheapest such container.

(a) Show that of all the rectangles with a given area, the one with smallest perimeter is a square. (b) Show that of all the rectangles with a given perimeter, the one with greatest area is a square.

Solve the initial-value problem. \(\frac{d P}{d t}=2 e^{3 t}, \quad t \geqslant 0, \quad P(0)=1\)

$$ \begin{array}{l}{37-44} \\ {\text { (a) Find the vertical and horizontal asymptotes. }} \\ {\text { (b) Find the intervals of increase or decrease. }} \\\ {\text { (c) Find the local maximum and minimum values. }} \\ {\text { (d) Find the intervals of concavity and the inflection points. }} \\ {\text { (e) Use the information from parts (a)-(d) to sketch the graph }} \\ {\text { of } f .}\end{array} $$ $$ f(x)=\frac{x^{2}}{x^{2}-1} $$

Find \(f\) . \(f^{\prime}(x)=2 x-3 / x^{4}, \quad x>0, \quad f(1)=3\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

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.