/*! 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 5 \(1-8\) " Solve the differential... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(1-8\) " Solve the differential equation. $$(y+\sin y) y^{\prime}=x+x^{3}$$

Short Answer

Expert verified
Equation involves intricate functions, likely needing substitution beyond typical by-hand resolution.

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \((y + \sin y) y' = x + x^3\). This is a first-order equation because it involves the first derivative \(y'\). Our task is to determine whether this is separable or another form we can solve.
02

Check for Separable Form

To determine if the equation is separable, we should see if we can write it as a product of a function of \(y\) and a function of \(x\). Rewriting the equation: \( y' = \frac{x + x^3}{y + \sin y} \), we notice that the terms are not easily separable as they multiply functions of \(y\) and \(x\).
03

Employ an Integrating Factor or Substitution

Given the complexity of separation, let's consider a substitution or integrating factor. Note: a first look suggests substitution might help if we let \(u = y + \sin y\). Thus, \(du = (1 + \cos y) dy = y' dx\), ensuring \(y'\) involves terms from both substitutions.
04

Substitute and Simplify

Substitute \(u = y + \sin y\) into the differential equation. Notice that \(y' = \frac{du}{dx} = \frac{1}{1+\cos y} (x + x^3)\) simplifies the problem domain. Evaluating this does not straightforwardly help, implying a trick or numeric solve might help further.
05

Evaluate for Solution Methods

Reassessing for either an integrable method or numeric solves within substitutions. Simplifying and trial multipliers or functions (from known library methods like trigonometric identities) might expedite handling both sides to corependency. Example check leads to lateral exploration.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

first-order differential equation
A first-order differential equation involves the first derivative of a function. In the given exercise, the equation \( (y + \sin y) y' = x + x^3 \) is a classic example of this type. Here, you see the derivative \( y' \), which denotes the rate of change of \( y \) concerning \( x \).

First-order differential equations are fundamental in many scientific fields, modeling things like population growth, heat transfer, and motion.

To understand them, it's crucial to recognize what a first derivative represents – essentially, it describes how \( y \) (our dependent variable) changes as \( x \) (the independent variable) changes. This is why first-order equations are commonly used to describe dynamical systems where change is inherent in the system itself.
separable differential equations
Separable differential equations are special because they allow us to separate variables and integrate them independently. However, not every differential equation is separable. Let's look at how to recognize and solve separable equations effectively.

For our given equation, \( y' = \frac{x + x^3}{y + \sin y} \), we attempted to check if it could be restructured into a form where one side involves only \( y \) and \( y' \) and the other only \( x \). Unfortunately, in this case, the equation doesn't easily allow for such separation.

Separable equations are beneficial because once separated, they can be integrated directly. This separation often involves rewriting the equation in the form:
  • one side as a function of \( y \) and \( dy \)
  • the other as a function of \( x \) and \( dx \)
When this is possible, you can integrate both sides with respect to their variables, leading to a solution.
integrating factor
When a differential equation does not lend itself to separation, we often use an integrating factor. An integrating factor is a function that, when multiplied through an equation, allows it to become integrable. This transforms the differential equation into a form where integration becomes possible.

In our exercise, though we initially consider substitution as a method, integrating factor techniques remain a mainstay for non-separable first-order linear differential equations. If applicable, it enables the left side of the equation to match the derivative of a product, simplifying to:
  • \((\mu(x) y)' \) on one side
  • \(\mu(x) Q(x) \) on the other
Here, \( \mu(x) \) is our integrating factor, often determined by a specific formula or known method for linear equations. While this wasn't the direct method for our problem, an understanding of it primes us for effectively tackling cases where it is exactly needed even in related trigonometric scenarios.

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

Sketch the region bounded by the curves, and visually estimate the location of the centroid. Then find the exact coordinates of the centroid. \(y=\sin x, \quad y=0, \quad 0 \leqslant x \leqslant \pi\)

The table shows values of a force function \(f(x),\) where \(x\) is measured in meters and \(f(x)\) in newtons. Use the Midpoint Rule to estimate the work done by the force in moving an object from \(x=4\) to \(x=20\) . $$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline x & {4} & {6} & {8} & {10} & {12} & {14} & {16} & {18} & {20} \\ \hline f(x) & {5} & {5.8} & {7.0} & {8.8} & {9.6} & {8.2} & {6.7} & {5.2} & {4.1} \\ \hline\end{array}$$

The curves with equations \(x^{n}+y^{n}=1, n=4,6,8, \ldots\) are called fat circles. Graph the curves with \(n=2,4,6,8,\) and 10 to see why. Set up an integral for the length \(L_{2 k}\) of the fat circle with \(n=2 k .\) Without attempting to evaluate this integral, state the value of \(\lim _{k \rightarrow \infty} L_{2 k}\) .

Sketch the region bounded by the curves, and visually estimate the location of the centroid. Then find the exact coordinates of the centroid. \(y=e^{x}, \quad y=0, \quad x=0, \quad x=1\)

An object of mass \(m\) is moving horizontally through a medium which resists the motion with a force that is a function of the velocity; that is, $$m \frac{d^{2} s}{d t^{2}}=m \frac{d v}{d t}=f(v)$$ where \(v=v(t)\) and \(s=s(t)\) represent the velocity and position of the object at time \(t,\) respectively. For example, think of a boat moving through the water. (a) Suppose that the resisting force is proportional to the velocity, that is, \(f(v)=-k v, k\) a positive constant. (This model is appropriate for small values of v.) Let \(v(0)=v_{0}\) and \(s(0)=s_{0}\) be the initial values of \(v\) and \(s\) Determine \(v\) and \(s\) at any time \(t .\) What is the total distance that the object travels from time \(t=0 ?\) (b) For larger values of \(v\) a better model is obtained by supposing that the resisting force is proportional to the square of the velocity, that is, \(f(v)=-k v^{2}, k>0 .\) (This model was first proposed by Newton.) Let \(v_{0}\) and \(s_{0}\) be the initial values of \(v\) and \(s .\) Determine \(v\) and \(s\) at any time \(t .\) What is the total distance that the object travels in this case?

See all solutions

Recommended explanations on Math Textbooks

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.