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Chapter 2: Problem 18

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

Short Answer

Expert verified
The solution to the differential equation is \( y = \frac{1}{-1 + \frac{1}{x} + \frac{C}{x e^x}} \).

Step by step solution

01

Substitute and Transform

The substitution given is \( w = y^{-1} \). Using the chain rule, \( \frac{dw}{dx} = -y^{-2} \frac{dy}{dx} \) which transforms the original equation into an equation in terms of \( w \). This gives us \( \frac{dw}{dx} + \left(1 + \frac{1}{x}\right) w = -1 \).
02

Identify Integrating Factor

To solve the linear first-order differential equation \( \frac{dw}{dx} + \left(1+\frac{1}{x}\right) w = -1 \), identify the integrating factor. The integrating factor here is given as \( x e^x \).
03

Apply Integrating Factor

Multiply the entire differential equation \( \frac{dw}{dx} + \left(1+\frac{1}{x}\right)w = -1 \) by the integrating factor \( x e^x \). This yields \( x e^x \frac{dw}{dx} + x e^x \left(1+\frac{1}{x}\right)w = -x e^x \).
04

Form a Perfect Derivative

The left-hand side becomes a perfect derivative: \( \frac{d}{dx}(x e^x w) = -x e^x \). This allows integration on both sides with respect to \( x \).
05

Integrate Both Sides

Integrate the equation \( \frac{d}{dx}(x e^x w) = -x e^x \) with respect to \( x \). The left side integrates to \( x e^x w \), and the right side integrates to \( -x e^x + e^x + C \), where \( C \) is a constant of integration.
06

Solve for \( w \)

We have \( x e^x w = -x e^x + e^x + C \). Solve for \( w \) to obtain \( w = -1 + \frac{1}{x} + \frac{C}{x e^x} \).
07

Substitute Back to \( y \)

Recall that \( w = y^{-1} \), so \( y^{-1} = -1 + \frac{1}{x} + \frac{C}{x e^x} \). Solving for \( y \) gives \( y = \frac{1}{-1 + \frac{1}{x} + \frac{C}{x e^x}} \).

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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 powerful tool used to solve linear first-order differential equations. It is a specially chosen function that, when multiplied by the original equation, transforms it into a form that is simpler to integrate. This makes it easier to find a solution to the differential equation.

In the given exercise, the differential equation is \[ \frac{dw}{dx} + \left(1 + \frac{1}{x}\right)w = -1 \]The integrating factor has been identified as \( xe^x \). By multiplying the entire equation by this integrating factor, the left side becomes a perfect derivative, simplifying the integration process.

  • Multiply each term in the equation by the integrating factor \( xe^x \).
  • This transforms the equation to \( xe^x \frac{dw}{dx} + xe^x \left(1 + \frac{1}{x}\right)w = -xe^x \).
  • The left-hand side becomes \( \frac{d}{dx}(xe^x w) \), which is a perfect derivative.
After applying the integrating factor, the equation is ready for the next step, integration.
Linear Differential Equation
A linear differential equation is one where the function and its derivatives appear to the first power, and there are no products of the function with its derivatives. It typically takes the form:\[\frac{dy}{dx} + P(x)y = Q(x)\]

For our exercise, after substitution and transformation, we are dealing with the linear differential equation:\[ \frac{dw}{dx} + \left(1 + \frac{1}{x}\right)w = -1 \]Here, \( P(x) = 1 + \frac{1}{x} \) and \( Q(x) = -1 \). Linear equations are crucial because they are easier to solve than nonlinear ones. By identifying such a structure, we can apply methods like finding an integrating factor.

  • Recognizing the linear form allows us to employ systematic methods such as using an integrating factor.
  • This characteristic simplifies the process significantly as we transform and integrate the equation accordingly.
Understanding the nature of linear differential equations is fundamental in finding practical solutions efficiently.
Variable Substitution
Variable substitution is an essential method in solving differential equations as it simplifies complex equations into more manageable forms. In this exercise, the substitution \( w = y^{-1} \) was used, which allowed the original equation involving \( y \) to be rewritten in terms of \( w \).

Here's how substitution works in this context:
  • Start with the original equation: \( y' - \left(1 + \frac{1}{x}\right) y = y^2 \).
  • Substitute \( w = y^{-1} \) and use the chain rule to express \( \frac{dw}{dx} \).
  • This makes the equation: \( \frac{dw}{dx} + \left(1 + \frac{1}{x}\right) w = -1 \).
By performing the substitution, the resulting equation becomes linear and more approachable. This strategy is a cornerstone in solving differential equations, enabling us to use standard methods more effectively.

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Most popular questions from this chapter

Writing the differential equation as \(\frac{d E}{d t}+\frac{1}{R C} E=0\) we see that an integrating factor is \(e^{t / R C}\). Then $$\begin{aligned} \frac{d}{d t}\left[e^{t / R C} E\right] &=0 \\ e^{t / R C} E &=c \\ E &=c e^{-t / R C} \end{aligned}$$ From \(E(4)=c e^{-4 / R C}=E_{0}\) we find \(c=E_{0} e^{4 / R C} .\) Thus, the solution of the initial-value problem is $$E=E_{0} e^{4 / R C} e^{-t / R C}=E_{0} e^{-(t-4) / R C}$$

For \(y^{\prime}+\frac{1}{x} y=\frac{1}{x} e^{x}\) an integrating factor is \(e^{\int(1 / x) d x}=x\) so that \(\frac{d}{d x}[x y]=e^{x}\) and \(y=\frac{1}{x} e^{x}+\frac{c}{x}\) for \(0< x<\infty\). If \(y(1)=2\) then \(c=2-e\) and \(y=\frac{1}{x} e^{x}+\frac{2-e}{x}\).

We first note that \(s(t)+i(t)+r(t)=n .\) Now the rate of change of the number of susceptible persons, \(s(t)\) is proportional to the number of contacts between the number of people infected and the number who are susceptible; that is, \(d s / d t=-k_{1} s i .\) We use \(-k_{1}<0\) because \(s(t)\) is decreasing. Next, the rate of change of the number of persons who have recovered is proportional to the number infected; that is, \(d r / d t=k_{2} i\) where \(k_{2}>0\) since \(r\) is increasing. Finally, to obtain \(d i / d t\) we use \\[\frac{d}{d t}(s+i+r)=\frac{d}{d t} n=0\\] This gives \\[\frac{d i}{d t}=-\frac{d r}{d t}-\frac{d s}{d t}=-k_{2} i+k_{1} s i\\] The system of differential equations is then $$\begin{aligned}&\frac{d s}{d t}=-k_{1} s i\\\&\frac{d i}{d t}=-k_{2} i+k_{1} s i\\\&\frac{d r}{d t}=k_{2} i\end{aligned}$$ A reasonable set of initial conditions is \(i(0)=i_{0},\) the number of infected people at time \(0, s(0)=n-i_{0},\) and \(r(0)=0\).

The differential equation for the first container is \(d T_{1} / d t=k_{1}\left(T_{1}-0\right)=k_{1} T_{1},\) whose solution is \(T_{1}(t)=c_{1} e^{k_{1} t}\) since \(T_{1}(0)=100\) (the initial temperature of the metal bar), we have \(100=c_{1}\) and \(T_{1}(t)=100 e^{k_{1} t}\). After 1 minute, \(T_{1}(1)=100 e^{k_{1}}=90^{\circ} \mathrm{C},\) so \(k_{1}=\ln 0.9\) and \(T_{1}(t)=100 e^{t \ln 0.9} .\) After 2 minutes, \(T_{1}(2)=100 e^{2 \ln 0.9}=\) \(100(0.9)^{2}=81^{\circ} \mathrm{C}\). The differential equation for the second container is \(d T_{2} / d t=k_{2}\left(T_{2}-100\right),\) whose solution is \(T_{2}(t)=\) \(100+c_{2} e^{k_{2} t} .\) When the metal bar is immersed in the second container, its initial temperature is \(T_{2}(0)=81,\) so $$T_{2}(0)=100+c_{2} e^{k_{2}(0)}=100+c_{2}=81$$ and \(c_{2}=-19 .\) Thus, \(T_{2}(t)=100-19 e^{k_{2} t} .\) After 1 minute in the second tank, the temperature of the metal bar is \(91^{\circ} \mathrm{C},\) so $$\begin{aligned} T_{2}(1) &=100-19 e^{k_{2}}=91 \\ e^{k_{2}} &=\frac{9}{19} \\ k_{2} &=\ln \frac{9}{19} \end{aligned}$$ and \(T_{2}(t)=100-19 e^{t \ln (9 / 19)} .\) Setting \(T_{2}(t)=99.9\) we have $$\begin{aligned} 100-19 e^{t \ln (9 / 19)} &=99.9 \\ e^{t \ln (9 / 19)} &=\frac{0.1}{19} \\ t &=\frac{\ln (0.1 / 19)}{\ln (9 / 19)} \approx 7.02 \end{aligned}$$ Thus, from the start of the "double dipping" process, the total time until the bar reaches \(99.9^{\circ} \mathrm{C}\) in the second container is approximately 9.02 minutes.

From \(\frac{1}{1-2 y} d y=d t\) we obtain \(-\frac{1}{2} \ln |1-2 y|=t+c\) or \(1-2 y=c_{1} e^{-2 t} .\) Using \(y(0)=5 / 2\) we find \(c_{1}=-4\). The solution of the initial-value problem is \(1-2 y=-4 e^{-2 t}\) or \(y=2 e^{-2 t}+\frac{1}{2}\).
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