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Chapter 2: Problem 16

In Problems 9-16, solve the given differential equation. $$ \left(2 r^{2} \cos \theta \sin \theta+r \cos \theta\right) d \theta+\left(4 r+\sin \theta-2 r \cos ^{2} \theta\right) d r=0 $$

Short Answer

Expert verified
The solution is \( r^2 \sin^2 \theta + r \sin \theta + 2r^2 = C \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is written in the form \( M(r, \theta) \, d\theta + N(r, \theta) \, dr = 0 \), where \( M = 2r^2 \cos \theta \sin \theta + r \cos \theta \) and \( N = 4r + \sin \theta - 2r \cos^2 \theta \). It is not readily apparent if it's exact, so we will check if the differential equation is exact or needs any manipulation to solve.
02

Check for Exactness

To check for exactness, compute the partial derivatives: \( \frac{\partial M}{\partial r} \) and \( \frac{\partial N}{\partial \theta} \). If they are equal, the differential equation is exact.Calculate \( \frac{\partial M}{\partial r} = \frac{d}{dr}(2r^2 \cos \theta \sin \theta + r \cos \theta) = 4r \cos \theta \sin \theta + \cos \theta \).Calculate \( \frac{\partial N}{\partial \theta} = \frac{d}{d\theta}(4r + \sin \theta - 2r \cos^2 \theta) = \cos \theta + 4r \cos \theta \sin \theta \).Since \( \frac{\partial M}{\partial r} = \frac{\partial N}{\partial \theta} \), the differential equation is exact.
03

Solve the Exact Differential Equation

Since the equation is exact, there exists a potential function \( \Psi(r, \theta) \) such that \( \Psi_r = N \) and \( \Psi_\theta = M \). Find \( \Psi \) by integrating \( M \) with respect to \( \theta \) and \( N \) with respect to \( r \).First, integrate \( M(r, \theta) \) with respect to \( \theta \):\[ \int (2r^2 \cos \theta \sin \theta + r \cos \theta) \, d\theta = r^2 \sin^2 \theta + r \sin \theta + f(r) \].Now, find \( f(r) \) by using \( \Psi_r = N \):\( \frac{\partial}{\partial r}[r^2 \sin^2 \theta + r \sin \theta + f(r)] = 2r \sin^2 \theta + \sin \theta + f'(r) = N = 4r + \sin \theta - 2r \cos^2 \theta \).Cancel out terms to find \( f'(r) \):\( f'(r) = 4r - 2r \cos^2 \theta \).Integrate \( f'(r) \) to find \( f(r) \):\( f(r) = 2r^2 + C \), where \( C \) is a constant.Thus, the potential function is:\[ \Psi(r, \theta) = r^2 \sin^2 \theta + r \sin \theta + 2r^2 = constant. \]
04

Write the Solution

From the potential function \( \Psi(r, \theta) = r^2 \sin^2 \theta + r \sin \theta + 2r^2 \), we equate it to a constant to find the solution:\[ r^2 \sin^2 \theta + r \sin \theta + 2r^2 = C \].This equation gives the implicit solution of the original differential equation, describing the relationship between \( r \) and \( \theta \).

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Key Concepts

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

Differential Equations
Differential equations are mathematical expressions that relate the rates at which quantities change. These equations consist of functions and their derivatives. In our problem, the equation combines the variables \( r \) and \( \theta \) in a way that their rates of change are interrelated through their derivatives.
  • The given differential equation is written in terms of \( M(r, \theta) \, d\theta + N(r, \theta) \, dr = 0 \), meaning it involves both the differentials \( d\theta \) and \( dr \).
  • This type of equation suggests a relationship where the total differential sums to zero, a common form used in solving exact differential equations.
  • Our task within such a context is to first determine if the relationship is an exact differential equation, which is crucial to finding a potential function that can simplify the process of integration and solution finding.
Understanding the nature of differential equations and their forms enables us to apply appropriate solution methods effectively to analyze the underlying relationships between variables.
Partial Derivatives
Partial derivatives are used in multivariable calculus to find the rate of change of a function concerning one of several variables. In the context of our differential equation, they are key to checking for exactness.
  • To confirm if an equation is exact, compute the partial derivatives of \( M \) with respect to \( r \), denoted \( \frac{\partial M}{\partial r} \), and \( N \) with respect to \( \theta \), denoted \( \frac{\partial N}{\partial \theta} \).
  • If these partial derivatives match, it confirms that the equation is exact, implying the existence of a potential function \( \Psi(r, \theta) \).
  • In our example, this test proved true: \( \frac{\partial M}{\partial r} = 4r \cos \theta \sin \theta + \cos \theta \) and \( \frac{\partial N}{\partial \theta} = \cos \theta + 4r \cos \theta \sin \theta \).
Partial derivatives hence offer a crucial tool for diagnosing our differential equation, simplifying our task by confirming the exactness and leading us to a potential solution path.
Integration
Integration is the process of finding the integral of functions, which is essentially the reverse operation of differentiation. In solving exact differential equations, integration aids in deriving the potential function.
  • Once exactness is confirmed, integration is employed to find the potential function \( \Psi \) that encapsulates the solution to the differential equation.
  • The integral of \( M(r, \theta) \) with respect to \( \theta \) leads to part of \( \Psi \), for instance, \( \int (2r^2 \cos \theta \sin \theta + r \cos \theta) \, d\theta = r^2 \sin^2 \theta + r \sin \theta + f(r) \).
  • The function \( f(r) \) is then determined by integrating the derivative \( f'(r) \), founded on the equality \( \Psi_r = N \).
Thorough integration is pivotal for solving differential equations by structuring predominant functions in an accessible form, leading to a clearer path to solutions.
Potential Function
The potential function is an essential component when dealing with exact differential equations. It is a function \( \Psi(r, \theta) \) where its derivatives with respect to each variable match the components \( M \) and \( N \) of the differential equation.
  • To find the potential function, we use integration of the components and ensure consistency with both partial derivatives \( \Psi_r = N \) and \( \Psi_\theta = M \).
  • For the differential equation \( M\,d\theta + N\,dr = 0 \), the potential function enables one to understand the relationship between \( r \) and \( \theta \) concisely: \( \Psi(r, \theta) = r^2 \sin^2 \theta + r \sin \theta + 2r^2 \).
  • The equation \( \Psi(r, \theta) = C \), where \( C \) is a constant, becomes the solution to the differential equation, representing a level curve in the space of \( r \) and \( \theta \).
The potential function is a powerful concept, offering an elegant solution to complex differential equations by reducing them to simple geometric interpretations.

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Most popular questions from this chapter

When interest is compounded continuously, the amount of money increases at a rate proportional to the amount \(S\) present at time \(t\), that is, \(d S / d t=r S\), where \(r\) is the annual rate of interest. (a) Find the amount of money accrued at the end of 5 years when \(\$ 5000\) is deposited in a savings account drawing \(5 \frac{3}{4} \%\) annual interest compounded continuously. (b) In how many years will the initial sum deposited bave doubled? (c) Use a calculator to compare the amount obtained in part (a) with the amount \(S=5000\left(1+\frac{1}{4}(0.0575)\right)^{5(4)}\) that is arcrued when interest is compounded quarterly.

A dead body was found within a closed room of a bouse where the temperature was a constant \(70^{\circ} \mathrm{F}\). At the time of discovery, the core temperature of the body was determined to be \(85^{\circ} \mathrm{F}\). One hour later a second measurement showed that the core temperature of the body was \(80^{\circ} \mathrm{F}\). Assume that the time of death corresponds to \(t=0\) and that the core temperature at that time was \(98.6^{\circ} \mathrm{F}\). Determine how many hours elapsed before the body was found.

(a) In Examples 3 and 4 of Section 2.1, we saw that any solution \(P(t)\) of \((4)\) possesses the asymptotic behavior \(P(t) \rightarrow a / b\) as \(t \rightarrow \infty\) for \(P_{0}>a / b\) and for \(0

Solve the given initial-value problem. Give the largest interval \(I\) over which the solution is defined. $$ \left(\frac{e^{-2 \sqrt{x}}-y}{\sqrt{x}}\right) \frac{d x}{d y}=1, \quad y(1)=1 $$

(a) The sine integral function is defined by \(\operatorname{Si}(x)=\) \(\int_{0}^{x}(\sin t / t) d t\), where the integrand is defined to be 1 at \(t=0\). Express the solution \(y(x)\) of the initial-value problem \(x^{3} y^{\prime}+2 x^{2} y=10 \sin x, y(1)=0\), in terms of \(\operatorname{Si}(x)\). (b) Use a CAS to graph the solution curve for the IVP for \(x>0 .\) (c) Use a CAS to find the value of the absolute maximum of the solution \(y(x)\) for \(x>0\).
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