/*! 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 11 First find the general solution ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 11

First find the general solution (involving a constant C) for the given differential equation. Then find the particular solution that satisfies the indicated condition. (See Example 2.) \(\frac{d s}{d t}=16 t^{2}+4 t-1 ; s=100\) at \(t=0\)

Short Answer

Expert verified
Particular solution is \(s(t) = \frac{16}{3}t^3 + 2t^2 - t + 100\).

Step by step solution

01

Integrate the Differential Equation

To find the general solution, we need to integrate the differential equation \( \frac{ds}{dt} = 16t^2 + 4t - 1 \). This yields:\[s(t) = \int (16t^2 + 4t - 1)\, dt = \frac{16}{3}t^3 + \frac{4}{2}t^2 - t + C = \frac{16}{3}t^3 + 2t^2 - t + C.\]
02

Apply the Initial Condition

We are given the condition \( s(0) = 100 \). Substitute \( t = 0 \) and \( s = 100 \) into the general solution to find \( C \):\[100 = \frac{16}{3}(0)^3 + 2(0)^2 - 0 + C = C.\]Thus, \( C = 100 \).
03

Write the Particular Solution

Substitute \( C = 100 \) back into the general solution. The particular solution is:\[s(t) = \frac{16}{3}t^3 + 2t^2 - t + 100.\]

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 describe how a quantity changes in relation to another. In this exercise, we work with the differential equation \( \frac{ds}{dt} = 16t^2 + 4t - 1 \). This tells us how the function \( s(t) \), which depends on \( t \), changes as \( t \) changes.
These equations are crucial in many fields, like physics and engineering, as they model real-world systems:
  • Motion of objects
  • Heat transfer
  • Population growth
Understanding these equations allows us to predict how systems behave over time.
Integration
Integration is a fundamental concept in calculus used to reverse-differentiate a function. You can think of it as finding the area under a curve. In the context of our differential equation, integration helps us transition from \( \frac{ds}{dt} \) back to \( s(t) \).
When we integrate \( 16t^2 + 4t - 1 \), it involves finding antiderivatives:
  • \( \frac{16}{3}t^3 \) from \( 16t^2 \)
  • \( 2t^2 \) from \( 4t \)
  • \(-t \) from \(-1 \)
This integration results in the general solution \( s(t) = \frac{16}{3}t^3 + 2t^2 - t + C \), where \( C \) is an integration constant. This constant appears because integration can produce multiple families of solutions that differ by a constant term.
Initial Conditions
Initial conditions are specific conditions or values provided to solve for constant terms in differential equations. They help us find a unique solution for our problem.
For this exercise, we're given that \( s(0) = 100 \). This means when \( t = 0 \), the function \( s(t) \) should equal 100. By substituting these values into our general solution \( s(t) = \frac{16}{3}t^3 + 2t^2 - t + C \), we can solve for \( C \).
This process simplifies our results to a single function, aligning with the specific criteria from the initial conditions. Initial conditions are pivotal in real-life scenarios, ensuring models align with known data.
Particular Solution
The particular solution is the result we get when we apply specific initial conditions to the general solution of a differential equation. Our general solution is \( s(t) = \frac{16}{3}t^3 + 2t^2 - t + C \), and from our exercise, we found \( C = 100 \) by using the initial condition. Thus, the particular solution is \( s(t) = \frac{16}{3}t^3 + 2t^2 - t + 100 \).
In essence, the particular solution tailors the general equation to a specific scenario and ensures the model reflects real-world situations accurately.
This individualized solution holds the power to accurately capture the dynamics and behaviors specified by the original problem, offering precise predictions.

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

In installment buying, one would like to figure out the real interest rate (effective rate), but unfortunately this involves solving a complicated equation. If one buys an item worth $$\$ P$$ today and agrees to pay for it with payments of $$\$ R$$ at the end of each month for \(k\) months, then $$ P=\frac{R}{i}\left[1-\frac{1}{(1+i)^{k}}\right] $$ where \(i\) is the interest rate per month. Tom bought a used car for $$\$ 2000$$ and agreed to pay for it with $$\$ 100$$ payments at the end of each of the next 24 months. (a) Show that \(i\) satisfies the equation \(20 i(1+i)^{24}-(1+i)^{24}+1=0\) (b) Show that Newton's Method for this equation reduces to $$ i_{n+1}=i_{n}-\left[\frac{20 i_{n}^{2}+19 i_{n}-1+\left(1+i_{n}\right)^{-23}}{500 i_{n}-4}\right] $$ (c) Find \(i\) accurate to five decimal places starting with \(i=0.012\), and then give the annual rate \(r\) as a percent \((r=1200 i)\).

In Problems 5-14, use Newton's Method to approximate the indicated root of the given equation accurate to five decimal places. Begin by sketching a graph. The positive root of \(2 x^{2}-\sin x=0\)

Using the same axes, draw the graphs for \(0 \leq t \leq 100\) of the following two models for the growth of world population (both described in this section). (a) Exponential growth: \(y=6.4 e^{0.0132 t}\) (b) Logistic growth: \(y=102.4 /\left(6+10 e^{-0.030 t}\right)\) Compare what the two models predict for world population in 2010,2040 , and 2090 . Note: Both models assume that world population was \(6.4\) billion in \(2004(t=0)\).

\text { Find } \int f^{\prime \prime}(x) d x \text { if } f(x)=x \sqrt{x^{3}+1}

$$ \int(x+1)^{2} d x $$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

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.