/*! 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 84 Solve the initial value problems... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 84

Solve the initial value problems in Exercises 83 and 84. $$\frac{d^{2} y}{d x^{2}}=\sec ^{2} x, \quad y(0)=0 \quad \text { and } \quad y^{\prime}(0)=1$$

Short Answer

Expert verified
The solution is \( y = -\ln |\cos x| + x \).

Step by step solution

01

Integration of the Second Derivative

The given problem is as follows: \( \frac{d^2 y}{dx^2} = \sec^2 x \). To find \( y \), we first need to integrate \( \frac{d^2 y}{dx^2} \) to obtain \( \frac{dy}{dx} \). Integrating \( \sec^2 x \) with respect to \( x \), we get \( \frac{dy}{dx} = \tan x + C_1 \), where \( C_1 \) is the integration constant.
02

Solve for the First Integration Constant

Given the initial condition \( y'(0) = 1 \), substitute \( x = 0 \) and \( \frac{dy}{dx} = 1 \) into the equation \( \frac{dy}{dx} = \tan x + C_1 \). This gives \( 1 = \tan(0) + C_1 = 0 + C_1 \), so \( C_1 = 1 \).
03

Integration of the First Derivative

After finding \( C_1 \), our updated equation is \( \frac{dy}{dx} = \tan x + 1 \). Integrate \( \tan x + 1 \) to get \( y(x) \). The integral of \( \tan x \) is \( -\ln |\cos x| \) and the integral of \( 1 \) is \( x \). So, \( y = -\ln |\cos x| + x + C_2 \), where \( C_2 \) is another integration constant.
04

Solve for the Second Integration Constant

Given the initial condition \( y(0) = 0 \), substitute \( x = 0 \) and \( y = 0 \) into the equation \( y = -\ln |\cos x| + x + C_2 \). This results in \( 0 = -\ln |\cos(0)| + 0 + C_2 = -\ln(1) + C_2 = 0 + C_2 \), so \( C_2 = 0 \).
05

Final Solution

With \( C_1 = 1 \) and \( C_2 = 0 \), the solution to the differential equation is \( y = -\ln |\cos x| + x \).

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.

Initial Value Problem
An initial value problem involves solving a differential equation while also satisfying specific conditions at the beginning of the process, known as initial conditions. In this exercise, the differential equation is solved with two initial conditions:
  • \( y(0) = 0 \) indicates that when \( x = 0 \), the function \( y \) should equal 0.
  • \( y'(0) = 1 \) states that the derivative of \( y \) with respect to \( x \) should be 1 when \( x = 0 \).
These conditions are vital as they allow us to find the exact values of the integration constants, which in turn specify the unique solution to the differential equation. Initial values provide a snapshot of the function's behavior at a particular point, ensuring the solution curves accurately through that point.
Integration of Trigonometric Functions
Integration of trigonometric functions is crucial when dealing with differential equations involving trigonometric terms. In this exercise, the function \( \sec^2 x \) is integrated. Here's why it's important:
  • Each trigonometric function integrates differently. Recognizing these integrations is essential for solving differential equations.
  • Integration of \( \sec^2 x \) gives \( \tan x + C_1 \), where \( C_1 \) is the integration constant. This is because the derivative of \( \tan x \) is \( \sec^2 x \).
  • This process involves finding the antiderivative of the function to step from the second derivative to the first derivative.
Understanding the integration of trigonometric functions helps in identifying the relationships between derivatives and their original functions, thus aiding in solving these equations accurately.
Second Order Differential Equations
A second order differential equation, like the one in this problem, involves the second derivative, which is denoted by \( \frac{d^2 y}{dx^2} \). Here's how you tackle them:
  • Begin by integrating the second derivative to get the first derivative. This is done in the first step of the solution where the integral of \( \sec^2 x \) results in \( \tan x + C_1 \).
  • Integrate the first derivative to recover the original function \( y(x) \). In this problem, the integration results in \( -\ln |\cos x| + x + C_2 \).
  • Each integration introduces an arbitrary constant, which must be determined using the initial values provided.
The key to solving second order differential equations lies in understanding how each level of derivative relates to the original function through these integrations. The complexity arises because each solution step depends on applying calculus operations carefully and consistently to align with those initial conditions 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

Evaluate the integrals in Exercises \(43-66\) $$ \int_{0}^{\ln \sqrt{3}} \frac{e^{x} d x}{1+e^{2 x}} $$

In Exercises \(127-129\) , find the domain and range of each composite function. Then graph the composites on separate screens. Do the graphs make sense in each case? Give reasons for your answers. Comment on any differences you see. $$ \text { a. } y=\sin ^{-1}(\sin x) \quad \text { b. } y=\sin \left(\sin ^{-1} x\right) $$

The continuous extension of \((\sin x)^{x}\) to \([0, \pi]\) \begin{equation} \begin{array}{l}{\text { a. Graph } f(x)=(\sin x)^{x} \text { on the interval } 0 \leq x \leq \pi . \text { What }} \\\\\quad {\text { value would you assign to } f \text { to make it continuous at } x=0 ?} \\ {\text { b. Verify your conclusion in part (a) by finding } \lim _{x \rightarrow 0^{\prime}} f(x)} \\\\\quad {\text { with I'Hopital's Rule. }}\\\\{\text { c. Returning to the graph, estimate the maximum value of } f \text { on }} \\ \quad{[0, \pi] \text { . About where is max } f \text { taken on? }}\\\\{\text { d. Sharpen your estimate in part (c) by graphing } f^{\prime} \text { in the same }} \\\ \quad{\text { window to see where its graph crosses the } x \text { -axis. To simplify }} \\\\\quad {\text { your work, you might want to delete the exponential factor from }} \\\\\quad {\text { the expression for } f^{\prime} \text { and graph just the factor that has a zero. }}\end{array} \end{equation}

Evaluate the integrals in Exercises \(41-60\) $$\int_{-\pi / 4}^{\pi / 4} \cosh (\tan \theta) \sec ^{2} \theta d \theta$$

Find the limits in Exercises \(91-98\) $$\lim _{x \rightarrow 0} \frac{\sin ^{-1} 5 x}{x}$$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

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.