/*! 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 18 Let \(M=2 y \sin x \cos x-y+2 y^... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 18

Let \(M=2 y \sin x \cos x-y+2 y^{2} e^{x y^{2}}\) and \(N=-x+\sin ^{2} x+4 x y e^{x y^{2}}\) so that $$M_{y}=2 \sin x \cos x-1+4 x y^{3} e^{x y^{2}}+4 y e^{x y^{2}}=N_{x}$$ From \(f_{x}=2 y \sin x \cos x-y+2 y^{2} e^{x y^{2}}\) we obtain \(f=y \sin ^{2} x-x y+2 e^{x y^{2}}+h(y), h^{\prime}(y)=0,\) and \(h(y)=0\) solution is \(y \sin ^{2} x-x y+2 e^{x y^{2}}=c\)

Short Answer

Expert verified
The solution is \( y \sin^2 x - xy + 2 e^{xy^2} = c \).

Step by step solution

01

Verify the Condition for Exactness

To determine if the given differential equation is exact, we need to verify that \( M_y = N_x \). This can be checked by differentiating \( M \) with respect to \( y \) and \( N \) with respect to \( x \), given: \( M_y = 2 \sin x \cos x - 1 + 4xy^3 e^{xy^2} + 4y e^{xy^2} \). Similarly, \( N_x = 2 \sin x \cos x - 1 + 4xy^3 e^{xy^2} + 4y e^{xy^2} \). Since \( M_y = N_x \), the differential equation is exact.
02

Identify the Potential Function

Given the exact differential equation, identify the potential function \( f(x, y) \) such that \( f_x = M \) and \( f_y = N \). Start integrating \( M \) with respect to \( x \) to obtain the partial potential function: \( f = \int (2y \sin x \cos x - y + 2y^2 e^{xy^2})dx = y \sin^2 x - xy + 2 e^{xy^2} + h(y) \), where \( h(y) \) is an arbitrary function of \( y \).
03

Differentiate and Solve for h(y)

Differentiate the potential function found in Step 2 with respect to \( y \) to check what \( h'(y) \) should be for the potential equation to satisfy \( f_y = N \). From \( N = -x + \sin^2 x + 4xy e^{xy^2} \), set the partial derivative equal to \( f_y \) and solve: \( f_y = \sin^2 x - x + 4xy e^{xy^2} + h'(y) = -x + \sin^2 x + 4xy e^{xy^2} \). This simplifies to \( h'(y) = 0 \), hence \( h(y) = 0 \).
04

Write the General Solution

Now that \( h(y) \) is determined, the potential function \( f(x, y) = y \sin^2 x - xy + 2 e^{xy^2} \) can be used to write the solution to the differential equation. The general solution is \( f(x, y) = c \), or \( y \sin^2 x - xy + 2 e^{xy^2} = c \), where \( c \) is a constant.

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.

Potential Function
In the context of exact differential equations, a potential function acts as a bridge that connects the different partial derivatives and helps solve the equation. When you have an exact differential equation, there is a function, often denoted as \( f(x, y) \), whose total differential equals the original equation.The purpose of identifying a potential function is to simplify and solve the differential equation by reducing it to a recognizable form. For example, if given functions \( M \) and \( N \), constructing the potential function involves:
  • Integrating \( M \) with respect to \( x \)
  • Adding a function of \( y \), usually denoted as \( h(y) \), to incorporate terms related only to \( y \)
Once the potential function \( f(x, y) \) is determined, the solution to the differential equation can be expressed simply as \( f(x, y) = c \), where \( c \) is a constant.
Exactness Condition
The exactness condition is the criterion used to determine whether a differential equation is exact. It states that for a differential equation expressed as:\[ M(x, y)\,dx + N(x, y)\,dy = 0 \]to be exact, the partial derivative of \( M \) with respect to \( y \) must equal the partial derivative of \( N \) with respect to \( x \).Mathematically, this condition is represented as:\[ M_y = N_x \]Ensuring the equality of these partial derivatives means that the given differential equation can be represented as the total differential of some potential function \( f(x, y) \). This check assures that the equation can be solved using integration to find this potential function.
Differential Equation Solution
Once you have verified exactness and determined the potential function, you can find the differential equation's solution. The solution process involves using the potential function \( f(x, y) \) derived from previous steps.The general solution can be written as:
  • \( f(x, y) = c \)
Here, \( c \) is an arbitrary constant. Substituting back the potential function provides a form that highlights the interdependent relationship of \( x \) and \( y \) according to the original differential equation.
Integration of Partial Derivatives
Integration of partial derivatives is a crucial part of solving exact differential equations. It allows you to reconstruct the potential function from the given differential equation components.To integrate partial derivatives properly, follow these general steps:
  • Start by integrating \( M \) with respect to \( x \) to obtain a preliminary potential function. This step accounts for the \( x \)-dependent part of the potential function.
  • Don't forget to add \( h(y) \), a function of \( y \), since integration in terms of \( x \) may miss terms involving only \( y \).
  • Differentiate this integrated function with respect to \( y \) to determine any missing terms and compare with \( N \).
  • Finally, solve for \( h(y) \) by setting the differential of the preliminary function equal to \( N \), ensuring that all terms align correctly.
Careful integration and understanding the relationship between the partial derivatives ensure that the potential function satisfies the original differential equation.

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

For \(y^{\prime}+\frac{2}{x^{2}-1} y=\frac{x+1}{x-1}\) an integrating factor is \(e^{\int\left[2 /\left(x^{2}-1\right)\right] d x}=\frac{x-1}{x+1}\) so that \(\frac{d}{d x}\left[\frac{x-1}{x+1} y\right]=1\) and \((x-1) y=x(x+1)+c(x+1)\) for \(-1< x<1\).

Using separation of variables to solve \(d T / d t=k\left(T-T_{m}\right)\) we get \(T(t)=T_{m}+c e^{k t} .\) Using \(T(0)=70\) we find \(c=70-T_{m},\) so \(T(t)=T_{m}+\left(70-T_{m}\right) e^{k t} .\) Using the given observations, we obtain $$\begin{array}{c} T\left(\frac{1}{2}\right)=T_{m}+\left(70-T_{m}\right) e^{k / 2}=110 \\ T(1)=T_{m}+\left(70-T_{m}\right) e^{k}=145 \end{array}$$ Then, from the first equation, \(e^{k / 2}=\left(110-T_{m}\right) /\left(70-T_{m}\right)\) and $$\begin{aligned} e^{k}=\left(e^{k / 2}\right)^{2}=\left(\frac{110-T_{m}}{70-T_{m}}\right)^{2} &=\frac{145-T_{m}}{70-T_{m}} \\ \frac{\left(110-T_{m}\right)^{2}}{70-T_{m}} &=145-T_{m} \\ 12100-220 T_{m}+T_{m}^{2} &=10150-250 T_{m}+T_{m}^{2} \\ T_{m} &=390 \end{aligned}$$ The temperature in the oven is \(390^{\circ}\).

The left-hand derivative of the function at \(x=1\) is \(1 / e\) and the right- hand derivative at \(x=1\) is \(1-1 / e\). Thus, \(y\) is not differentiable at \(x=1\).

For \(\frac{d r}{d \theta}+r \sec \theta=\cos \theta\) an integrating factor is \(e^{\int \sec \theta d \theta}=e^{\ln |\sec x+\tan x|}=\sec \theta+\tan \theta\) so that \(\frac{d}{d \theta}[(\sec \theta+\tan \theta) r]=1+\sin \theta\) and \((\sec \theta+\tan \theta) r=\theta-\cos \theta+c\) for \(-\pi / 2 <\theta < \pi / 2\).

Assuming that the air resistance is proportional to velocity and the positive direction is downward with \(s(0)=0\) the model for the velocity is \(m d v / d t=m g-k v .\) Using separation of variables to solve this differential equation, we obtain \(v(t)=m g / k+c e^{-k t / m} .\) Then, using \(v(0)=0,\) we get \(v(t)=(m g / k)\left(1-e^{-k t / m}\right)\) Letting \(k=0.5, m=(125+35) / 32=5,\) and \(g=32,\) we have \(v(t)=320\left(1-e^{-0.1 t}\right) .\) Integrating we find \(s(t)=320 t+3200 e^{-0.1 t}+c_{1} . \quad\) Solving \(s(0)=0\) for \(c_{1}\) we find \(c_{1}=-3200,\) therefore \(s(t)=\) \(320 t+3200 e^{-0.1 t}-3200 .\) At \(t=15,\) when the parachute opens, \(v(15)=248.598\) and \(s(15)=2314.02\) At this time the value of \(k\) changes to \(k=10\) and the new initial velocity is \(v_{0}=248.598 .\) With the parachute open, the skydiver's velocity is \(v_{p}(t)=m g / k+c_{2} e^{-k t / m},\) where \(t\) is reset to 0 when the parachute opens. Letting \(m=5, g=32,\) and \(k=10,\) this gives \(v_{p}(t)=16+c_{2} e^{-2 t} .\) From \(v(0)=248.598\) we find \(c_{2}=232.598\) so \(v_{p}(t)=16+232.598 e^{-2 t} .\) Integrating, we get \(s_{p}(t)=16 t-116.299 e^{-2 t}+c_{3} .\) Solving \(s_{p}(0)=0\) for \(c_{3}\) we find \(c_{3}=116.299,\) so \(s_{p}(t)=16 t-116.299 e^{-2 t}+116.299 .\) Twenty seconds after leaving the plane is five seconds after the parachute opens. The skydiver's velocity at this time is \(v_{p}(5)=16.0106 \mathrm{ft} / \mathrm{s}\) and she has fallen a total of \(s(15)+s_{p}(5)=2314.02+196.294=2510.31 \mathrm{ft} .\) Her terminal velocity is \(\lim _{t \rightarrow \infty} v_{p}(t)=16, \mathrm{s}\) she has very nearly reached her terminal velocity five seconds after the parachute opens. When the parachute opens, the distance to the ground is \(15,000-s(15)=15,000-2,314=12,686\) ft. Solving \(s_{p}(t)=12,686\) we get \(t=785.6 \mathrm{s}=13.1 \mathrm{min} .\) Thus, it will take her approximately 13.1 minutes to reach the ground after her parachute has opened and a total of \((785.6+15) / 60=13.34\) minutes after she exits the plane.
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Nuclear Physics

Read Explanation

Electrostatics

Read Explanation

Waves Physics

Read Explanation

Engineering Physics

Read Explanation

Conservation of Energy and Momentum

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.