/*! 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 45 We are looking for a function \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 45

We are looking for a function \(y(x)\) such that \\[y^{2}+\left(\frac{d y}{d x}\right)^{2}=1\\]. Using the positive square root gives \\[\frac{d y}{d x}=\sqrt{1-y^{2}} \Longrightarrow \frac{d y}{\sqrt{1-y^{2}}}=d x \Longrightarrow \sin ^{-1} y=x+c\\]. Note that when \(c=c_{1}=0\) and when \(c=c_{1}=\pi / 2\) we obtain the well known particular solutions \(y=\sin x\) \(y=-\sin x, y=\cos x,\) and \(y=-\cos x .\) Note also that \(y=1\) and \(y=-1\) are singular solutions.

Short Answer

Expert verified
The solutions are \(y = \sin x, y = -\sin x, y = \cos x, y = -\cos x, y = 1, y = -1\).

Step by step solution

01

Understand the Given Equation

We start with the differential equation \(y^{2} + \left( \frac{dy}{dx} \right)^{2} = 1\). This equation describes a relationship between the function \(y\) and its derivative \(\frac{dy}{dx}\). Our task is to find function \(y(x)\) such that the equation holds.
02

Differentiate with respect to x

Recognize that we can explore the form \(\frac{dy}{dx} = \sqrt{1-y^2} \) based on the constraint \(y^{2} + \left( \frac{dy}{dx} \right)^{2} = 1\). Notice that this equation resembles the Pythagorean identity and considers the range for a trigonometric function.
03

Separate the Variables

Recast the equation \(\frac{dy}{dx} = \sqrt{1-y^2}\) into a separable form: \(\frac{dy}{\sqrt{1-y^2}} = dx\). This allows us to integrate each side with respect to its variable.
04

Integrate Both Sides

Integrate the left-hand side \(\int \frac{dy}{\sqrt{1-y^2}}\) which is known to give \(\sin^{-1} y + C\). Integrate the right-hand side \(\int dx = x + K\). Set the constants equal: \(C = K = c\).
05

Solve for y(x)

By equating the integrated forms, we derive the equation \(\sin^{-1} y = x + c\). Solving this, we find \(y = \sin(x + c)\).
06

Explore Particular Solutions

Select certain values of \(c\) to find particular solutions: - If \(c = 0\), we have \(y = \sin x\).- If \(c = \pi/2\), we get \(y = \cos x\).- Similarly, \(-y = \sin x\) implies \(y = -\sin x\), and \(-y = \cos x\) implies \(y = -\cos x\).
07

Identify Singular Solutions

The equation \(y^2 + \left( \frac{dy}{dx} \right)^2 = 1\) also admits the constant solutions \(y = 1\) and \(y = -1\), since they satisfy the equation.

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.

Separable Differential Equations
A differential equation is termed separable if you can rearrange it such that all terms with one variable appear on one side of the equation, and all terms with the other variable appear on the opposite side. This makes them easier to solve with integration. In the given exercise, we start with the equation: \( \frac{dy}{dx} = \sqrt{1-y^2} \). By separating the variables, we rearrange it as \( \frac{dy}{\sqrt{1-y^2}} = dx \). This setup allows us to integrate each side separately, ultimately obtaining a general solution. Here are a few key steps to remember when working with separable differential equations:
  • Rearrange terms so that one side of the equation contains only terms involving the variable \(y\) and \(dy\), and the other side contains terms involving \(x\) and \(dx\).
  • Integrate both sides independently.
  • Combine the constants of integration into a single constant.
Thus, separable differential equations form a foundational method for solving many problems in calculus by leveraging the process of integration.
Pythagorean Identity
The Pythagorean identity is a crucial concept when dealing with trigonometric functions. It states that for any angle \( \theta \), \( \sin^2 \theta + \cos^2 \theta = 1 \). In our problem, the equation \( y^2 + \left( \frac{dy}{dx} \right)^2 = 1 \) resembles this identity. This resemblance gives us a clue that the solutions to the differential equation might involve trigonometric functions.
  • Notice how the equation mirrors the structure of \( \sin^2 \theta + \cos^2 \theta = 1 \).
  • This similarity suggests that solutions involving sine and cosine functions may satisfy the initial differential equation.
Recognizing these patterns helps in identifying the potential form of the solutions. Therefore, understanding the Pythagorean identity can provide insight into solving trigonometric differential equations.
Singular Solutions
In the world of differential equations, singular solutions are those solutions which satisfy the differential equation but cannot be obtained directly from the general solution. In the given problem, besides the solutions derived using the integration of the separable equation, there are particular constant solutions: \( y = 1 \) and \( y = -1 \). These solutions do not vary with \(x\) and satisfy the equation independently:
  • For \( y = 1 \), the derivative \( \frac{dy}{dx} = 0 \), thus \( 1 + 0^2 = 1 \), which satisfies the equation.
  • Similarly, for \( y = -1 \), we have \( (-1)^2 + 0^2 = 1 \).
Such solutions highlight that not every solution needs to be generated from the general form derived through integration, emphasizing the uniqueness and applicability of singular solutions in differential equations.
Particular Solutions
A particular solution is a specific instance of a general solution to a differential equation, obtained by assigning specific values to the constant of integration. In the exercise provided, by choosing different values for \(c\), we can derive various particular solutions.
  • If \( c = 0 \), the function \( y = \sin x \) is obtained.
  • When \( c = \frac{\pi}{2} \), \( y = \cos x \) arises as a solution.
  • Additionally, using \(-y\) yields \( y = -\sin x \) and \( y = -\cos x \).
By exploring the different values for \(c\), we generate different functions that all solve the differential equation given the initial condition \(c\). Particular solutions offer a precise solution tailored to specific conditions or constraints applied to the 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

From \(\frac{y}{\left(1+y^{2}\right)^{1 / 2}} d y=\frac{x}{\left(1+x^{2}\right)^{1 / 2}} d x\) we obtain \(\left(1+y^{2}\right)^{1 / 2}=\left(1+x^{2}\right)^{1 / 2}+c\).

From \(y^{\prime}+\frac{1}{x} y=\frac{1}{x} y^{-2}\) and \(w=y^{3}\) we obtain \(\frac{d w}{d x}+\frac{3}{x} w=\frac{3}{x} .\) An integrating factor is \(x^{3}\) so that \(x^{3} w=x^{3}+c\) or \(y^{3}=1+c x^{-3}\).

(a) From \(m d v / d t=m g-k v\) we obtain \(v=m g / k+c e^{-k t / m} .\) If \(v(0)=v_{0}\) then \(c=v_{0}-m g / k\) and the solution of the initial-value problem is $$v(t)=\frac{m g}{k}+\left(v_{0}-\frac{m g}{k}\right) e^{-k t / m}$$. (b) As \(t \rightarrow \infty\) the limiting velocity is \(m g / k\) (c) From \(d s / d t=v\) and \(s(0)=0\) we obtain $$s(t)=\frac{m g}{k} t-\frac{m}{k}\left(v_{0}-\frac{m g}{k}\right) e^{-k t / m}+\frac{m}{k}\left(v_{0}-\frac{m g}{k}\right)$$.

(a) Separating variables and integrating, we have \\[\left(-2 y+y^{2}\right) d y=\left(x-x^{2}\right) d x\\] and \\[-y^{2}+\frac{1}{3} y^{3}=\frac{1}{2} x^{2}-\frac{1}{3} x^{3}+c\\]. Using a CAS we show some contours of \\[f(x, y)=2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}\\]. The plots shown on \([-7,7] \times[-5,5]\) correspond to \(c\) -values of -450,-300,-200,-120,-60,-20,-10,-8.1,-5 \(-0.8,20,60,\) and 120. (b) The value of \(c\) corresponding to \(y(0)=\frac{3}{2}\) is \(f\left(0, \frac{3}{2}\right)=-\frac{27}{4}\) The portion of the graph between the dots corresponds to the solution curve satisfying the intial condition. To determine the interval of definition we find \(d y / d x\) for \\[2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-\frac{27}{4}\\]. Using implicit differentiation we get \(y^{\prime}=\left(x-x^{2}\right) /\left(y^{2}-2 y\right)\) which is infinite when \(y=0\) and \(y=2 .\) Letting \(y=0\) in \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-\frac{27}{4}\) and using a CAS to solve for \(x\) we get \(x=-1.13232 .\) Similarly, letting \(y=2,\) we find \(x=1.71299 .\) The largest interval of definition is approximately (-1.13232,1.71299). (c) The value of \(c\) corresponding to \(y(0)=-2\) is \(f(0,-2)=-40\) The portion of the graph to the right of the dot corresponds to the solution curve satisfying the initial condition. To determine the interval of definition we find \(d y / d x\) for \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-40\). Using implicit differentiation we get \(y^{\prime}=\left(x-x^{2}\right) /\left(y^{2}-2 y\right)\) which is infinite when \(y=0\) and \(y=2 .\) Letting \(y=0\) in \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-40\) and using a CAS to solve for \(x\) we get \(x=-2.29551 .\) The largest interval of definition is approximately \((-2.29551, \infty)\).

From \(\frac{1}{y^{2}-1} d y=\frac{1}{x^{2}-1} d x\) or \(\frac{1}{2}\left(\frac{1}{y-1}-\frac{1}{y+1}\right) d y=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right) d x\) we obtain \(\ln |y-1|-\ln |y+1|=\ln |x-1|-\ln |x+1|+\ln c\) or \(\frac{y-1}{y+1}=\frac{c(x-1)}{x+1} . \quad\) Using \(y(2)=2\) we find \(c=1 .\) A solution of the initial-value problem is \(\frac{y-1}{y+1}=\frac{x-1}{x+1}\) or \(y=x\).
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Atoms and Radioactivity

Read Explanation

Geometrical and Physical Optics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Translational Dynamics

Read Explanation

Electricity and Magnetism

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.