/*! 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 21 For \(\frac{d r}{d \theta}+r \se... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 21

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 \(-\pi / 2 <\theta < \pi / 2\).

Short Answer

Expert verified
The integrating factor is \( \sec \theta + \tan \theta \). The solution is \( r(\theta) = \frac{\theta - \cos \theta + C}{\sec \theta + \tan \theta} \).

Step by step solution

01

Identifying the Equation Type

The given differential equation is \( \frac{dr}{d\theta} + r \sec \theta = \cos \theta \). This equation resembles a first-order linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \). Here, \(P(\theta) = \sec \theta\) and \(Q(\theta) = \cos \theta\).
02

Finding the Integrating Factor

The integrating factor for a first-order linear differential equation is given by \( e^{\int P(\theta) \, d\theta} \). Therefore, the integrating factor here is \( e^{\int \sec \theta \, d\theta} \). We know that \( \int \sec \theta \, d\theta = \ln |\sec \theta + \tan \theta| \), which simplifies to the integrating factor \( \sec \theta + \tan \theta \).
03

Multiplying through by the Integrating Factor

Multiply the entire original differential equation by the integrating factor \( \sec \theta + \tan \theta \) to obtain:\[ (\sec \theta + \tan \theta) \frac{dr}{d\theta} + r(\sec \theta + \tan \theta)^2 = \cos \theta (\sec \theta + \tan \theta) \].This can be simplified to:\[ \frac{d}{d\theta}[(\sec \theta + \tan \theta) r] = 1 + \sin \theta \].
04

Solving the Differential Equation

We integrate both sides of the equation \( \frac{d}{d\theta}[(\sec \theta + \tan \theta) r] = 1 + \sin \theta \):\[ \int \frac{d}{d\theta}[(\sec \theta + \tan \theta) r] \, d\theta = \int (1 + \sin \theta) \, d\theta \]. The left side integrates to \( (\sec \theta + \tan \theta) r \) and the right side results in \( \theta - \cos \theta + C \), where \( C \) is a constant. Thus:\( (\sec \theta + \tan \theta) r = \theta - \cos \theta + C \).
05

General Solution

The solution to the differential equation is \( r(\theta) = \frac{\theta - \cos \theta + C}{\sec \theta + \tan \theta} \) for \(-\pi/2 < \theta < \pi/2\). This represents the family of solutions dependent upon the constant \( C \).

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Key Concepts

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

Integrating Factor
An integrating factor is a mathematical tool used primarily to simplify the process of solving first-order linear differential equations. The idea is to transform the differential equation into a form that is easier to solve by making it integrable. You typically use the formula, which incorporates an exponential function, to find this factor.

To obtain the integrating factor, we first identify the function \( P(\theta) \) from the standard form of a first-order linear differential equation \( \frac{dy}{dx} + P(x) y = Q(x) \). For our differential equation, \( P(\theta) \) is \( \sec \theta \). The integrating factor is then found using the formula:
  • Integrating Factor = \( e^{\int P(\theta) \, d\theta} \)
Having this equation, we integrate \( \int \sec \theta \, d\theta \), which gives \( \ln |\sec \theta + \tan \theta| \). The exponential of this result leads us to the integrating factor:
  • \( \sec \theta + \tan \theta \)
This factor is crucial as it allows us to rewrite the original differential equation in a way that makes it possible to solve through simple integration.
General Solution
The general solution of a differential equation represents the entire set of possible solutions and is dependent on a constant that arises from the integration process. This constant reflects the initial or boundary conditions that could be applied to the equation.

Once we have multiplied the original differential equation by the integrating factor \( \sec \theta + \tan \theta \), we transform it into:
  • \( \frac{d}{d\theta}[(\sec \theta + \tan \theta) r] = 1 + \sin \theta \)
Solving involves integrating both sides. The left-hand side integrates directly to
  • \( (\sec \theta + \tan \theta) r \)
The integral of \( 1 + \sin \theta \) gives us \( \theta - \cos \theta + C \). Therefore, the rearranged expression is:
  • \( (\sec \theta + \tan \theta) r = \theta - \cos \theta + C \)
The general solution is then isolated as:
  • \( r(\theta) = \frac{\theta - \cos \theta + C}{\sec \theta + \tan \theta} \)
This equation describes an infinite family of solutions, determined by different possible values of the constant \( C \). Adjusting \( C \) accommodates particular solutions specific to certain initial conditions.
Differential Equation Solution
Solving a differential equation involves finding a function or set of functions, which satisfy the equation for given conditions. The process of solving can offer insights into the behavior modeled by the equation, such as system dynamics, without having to perform an experiment.

For our differential equation, applying the integrating factor transformed it into a form that allowed us to integrate easily. The process that followed was
  • Multiplying through by the integrating factor \( \sec \theta + \tan \theta \)
  • Recognizing that the left side is the derivative of \( (\sec \theta + \tan \theta) r \)
  • Integrating both sides of the equation
What's important here is understanding why each step is taken. Each manipulation aids in simplifying an otherwise complex equation or suggests a path toward a solution by integration. Finally, after integrating, the general solution \( r(\theta) = \frac{\theta - \cos \theta + C}{\sec \theta + \tan \theta} \) is achieved.

Given specific conditions, like initial values or boundary criteria, we can plug these into our general solution to pinpoint an exact or particular solution. This process reiterates why each step, starting from determining the equation type, finding the integrating factor, to integrating, is essential in solving first-order linear differential equations effectively.

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Most popular questions from this chapter

(a) An implicit solution of the differential equation \((2 y+2) d y-\left(4 x^{3}+6 x\right) d x=0\) is \\[y^{2}+2 y-x^{4}-3 x^{2}+c=0\\]. The condition \(y(0)=-3\) implies that \(c=-3 .\) Therefore \(y^{2}+2 y-x^{4}-3 x^{2}-3=0\). (b) Using the quadratic formula we can solve for \(y\) in terms of \(x\): \\[y=\frac{-2 \pm \sqrt{4+4\left(x^{4}+3 x^{2}+3\right)}}{2}\\]. The explicit solution that satisfies the initial condition is then \\[y=-1-\sqrt{x^{4}+3 x^{3}+4}\\]. (c) From the graph of the function \(f(x)=x^{4}+3 x^{3}+4\) below we see that \(f(x) \leq 0\) on the approximate interval \(-2.8 \leq x \leq-1.3 .\) Thus the approximate domain of the function $$y=-1-\sqrt{x^{4}+3 x^{3}+4}=-1-\sqrt{f(x)}$$ is \(x \leq-2.8\) or \(x \geq-1.3 .\) The graph of this function is shown below. (d) Using the root finding capabilities of a CAS, the zeros of \(f\) are found to be -2.82202 and \(-1.3409 .\) The domain of definition of the solution \(y(x)\) is then \(x>-1.3409 .\) The equality has been removed since the derivative \(d y / d x\) does not exist at the points where \(f(x)=0 .\) The graph of the solution \(y=\phi(x)\) is given on the right.

(a) \(\operatorname{From} 2 W^{2}-W^{3}=W^{2}(2-W)=0\) we see that \(W=0\) and \(W=2\) are constant solutions. (b) Separating variables and using a CAS to integrate we get \\[ \frac{d W}{W \sqrt{4-2 W}}=d x \quad \text { and } \quad-\tanh ^{-1}\left(\frac{1}{2} \sqrt{4-2 W}\right)=x+c \\] Using the facts that the hyperbolic tangent is an odd function and \(1-\tanh ^{2} x=\operatorname{sech}^{2} x\) we have \\[ \begin{aligned} \frac{1}{2} \sqrt{4-2 W} &=\tanh (-x-c)=-\tanh (x+c) \\ \frac{1}{4}(4-2 W) &=\tanh ^{2}(x+c) \\ 1-\frac{1}{2} W &=\tanh ^{2}(x+c) \\ \frac{1}{2} W &=1-\tanh ^{2}(x+c)=\operatorname{sech}^{2}(x+c) \end{aligned} \\] Thus, \(W(x)=2 \operatorname{sech}^{2}(x+c)\) (c) Letting \(x=0\) and \(W=2\) we find that \(\operatorname{sech}^{2}(c)=1\) and \(c=0.\)

From \(d A / d t=0-A / 50\) we obtain \(A=c e^{-t / 50} .\) If \(A(0)=30\) then \(c=30\) and \(A=30 e^{-t / 50}\).

From \(y e^{y} d y=\left(e^{-x}+e^{-3 x}\right) d x\) we obtain \(y e^{y}-e^{y}+e^{-x}+\frac{1}{3} e^{-3 x}=c\).

Assume that \(d T / d t=k(T-100)\) so that \(T=100+c e^{k t} .\) If \(T(0)=20^{\circ}\) and \(T(1)=22^{\circ},\) then \(c=-80\) and \(k=\ln (39 / 40)\) so that \(T(t)=90^{\circ},\) which implies \(t=82.1\) seconds. If \(T(t)=98^{\circ}\) then \(t=145.7\) seconds.
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