/*! 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 12 Verify that the given function \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 12

Verify that the given function \(y\) is a solution of the initial value problem that follows it. $$y(t)=8 t^{6}-3 ; t y^{\prime}(t)-6 y(t)=18, y(1)=5$$

Short Answer

Expert verified
Question: Verify that the function \(y(t) = 8t^6 - 3\) is a solution to the initial value problem \(ty'(t) - 6y(t) = 18\) with initial condition \(y(1) = 5\). Answer: After finding the first derivative \(y'(t) = 48t^5\), substituting the function and its first derivative into the initial value problem, and simplifying the resulting equation, the equation holds true. Moreover, the initial condition \(y(1) = 5\) is satisfied by the given function. Therefore, we can verify that \(y(t) = 8t^6 - 3\) is a solution to the initial value problem.

Step by step solution

01

Find the first derivative of the given function

First, we need to find the first derivative of the given function. The given function is: $$y(t) = 8t^6 - 3$$ We will differentiate this function with respect to \(t\). Using the power rule, we get: $$y'(t) = 6 \cdot 8t^5 = 48t^5$$
02

Substitute the function and its first derivative into the initial value problem

Now that we found the first derivative, we can plug \(y(t)\) and \(y'(t)\) back into the initial value problem: $$ty'(t) - 6y(t) = 18$$ By replacing \(y(t)\) by \(8t^6 - 3\) and \(y'(t)\) by \(48t^5\), we have: $$t(48t^5) - 6(8t^6 - 3) = 18$$
03

Simplify and check if the equation holds true

Let's simplify the equation and see if it holds true: \begin{align*} 48t^6 - 48t^6 + 18 &= 18 \\ 18 &=18 \end{align*} Since both sides of the equation are equal, the given function does satisfy the initial value problem.
04

Check the initial condition

Finally, we need to check if the initial condition \(y(1) = 5\) is also satisfied by the given function. Plugging \(t = 1\) into the given function, we get: $$y(1) = 8(1)^6 - 3 = 8 - 3 = 5$$ Thus, the initial condition is also satisfied. Since the given function satisfies both the initial value problem and the initial condition, we can now conclude that \(y(t) = 8t^6 - 3\) is indeed a solution to the initial value problem.

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.

Differentiation
Differentiation is a fundamental concept in calculus that focuses on finding the derivative of a function. A derivative represents how a function changes as its input changes. In simpler terms, it's the mathematical way of understanding the rate of change or the slope of the function at any given point.
For example, if we take a function like \(y(t) = 8t^6 - 3\), differentiation allows us to find its derivative, \(y'(t)\). The derivative tells us how the value of \(y\) changes for a small change in \(t\).
To differentiate effectively, we use specific rules like the power rule, which helps us compute derivatives easily and accurately. Understanding differentiation is crucial because it not only helps verify solutions in problems like the initial value problem but is also essential for optimizing functions and modeling situations in physics, engineering, and economics.
Function Verification
Function verification involves checking whether a specified function indeed satisfies a particular condition or set of conditions. In the context of initial value problems, this means determining if our function is a valid solution to the given differential equation and initial conditions.
For the problem considered, \(y(t) = 8t^6 - 3\) must satisfy both the differential equation \(ty'(t) - 6y(t) = 18\) and the initial condition \(y(1) = 5\).
Verification involves substituting the function and its derivative into the equation to confirm equality. If both sides of the equation match, the function is indeed a valid solution. Moreover, the initial condition is verified by ensuring that substituting the initial condition into the function results in the provided initial value.
Power Rule
The power rule is a key technique in differentiation that simplifies the process of finding derivatives. This rule states that if you have a function of the form \(x^n\), its derivative is \(nx^{n-1}\).
This makes differentiating polynomial functions straightforward and quick. For instance, considering the term \(8t^6\) in our function, the power rule is applied to find the derivative. The exponent \(6\) is multiplied by the coefficient \(8\) to get \(48\), and the new exponent becomes \(5\) after subtracting one, resulting in \(48t^5\).
The power rule greatly reduces the complexity involved in manual calculations, helping you tackle problems more efficiently. It is essential for any calculus student to master this rule, as it's widely used in both academic exercises and real-world applications.

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 physiological model A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like a stirred tank. Suppose that the blood volume is a four-liter tank that initially has a zero concentration of a particular drug. At time \(t=0,\) an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of \(500 \mathrm{mg} / \mathrm{L} .\) The inflow rate is \(0.06 \mathrm{L} / \mathrm{min}\). Assume that the drug is quickly mixed thoroughly in the blood and that the volume of blood remains constant. a. Write an initial value problem that models the mass of the drug in the blood, for \(t \geq 0\) b. Solve the initial value problem and graph both the mass of the drug and the concentration of the drug. c. What is the steady-state mass of the drug in the blood? d. After how many minutes does the drug mass reach \(90 \%\) of its steady-state level?

Determine whether the following equations are separable. If so, solve the initial value problem. $$2 y y^{\prime}(t)=3 t^{2}, y(0)=9$$

A differential equation of the form \(y^{\prime}(t)=f(y)\) is said to be autonomous (the function \(f\) depends only on \(y\) ). The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(f\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations. a. Find the equilibrium solutions. b. Sketch the direction field, for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=y(2-y)$$

Determine whether the following equations are separable. If so, solve the initial value problem. $$y^{\prime}(t)=\cos ^{2} y, y(1)=\frac{\pi}{4}$$

When an infected person is introduced into a closed and otherwise healthy community, the number of people who contract the disease (in the absence of any intervention) may be modeled by the logistic equation $$ \frac{d P}{d t}=k P\left(1-\frac{P}{A}\right), P(0)=P_{0} $$ where \(k\) is a positive infection rate, \(A\) is the number of people in the community, and \(P_{0}\) is the number of infected people at \(t=0\) The model also assumes no recovery. a. Find the solution of the initial value problem, for \(t \geq 0\), in terms of \(k, A,\) and \(P_{0}\) b. Graph the solution in the case that \(k=0.025, A=300,\) and \(P_{0}=1\) c. For a fixed value of \(k\) and \(A\), describe the long-term behavior of the solutions, for any \(P_{0}\) with \(0 < P_{0} < A\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Probability and Statistics

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.