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Chapter 2: Problem 36

We note that \(\left(N_{x}-M_{y}\right) / M=-3 / y,\) so an integrating factor is \(e^{-3} \int d y / y=1 / y^{3} .\) Let \(M=\left(y^{2}+x y^{3}\right) / y^{3}=\) \(1 / y+x\) and \(N=\left(5 y^{2}-x y+y^{3} \sin y\right) / y^{3}=5 / y-x / y^{2}+\sin y,\) so that \(M_{y}=-1 / y^{2}=N_{x} .\) From \(f_{x}=1 / y+x\) we obtain \(f=x / y+\frac{1}{2} x^{2}+h(y), h^{\prime}(y)=5 / y+\sin y,\) and \(h(y)=5 \ln |y|-\cos y .\) A solution of the differential equation is \(x / y+\frac{1}{2} x^{2}+5 \ln |y|-\cos y=c\)

Short Answer

Expert verified
The solution is \(x/y + 1/2 \, x^2 + 5 \, \ln |y| - \cos y = c\).

Step by step solution

01

Identify Integrating Factor

The given equation is expressed in terms of the integrating factor. We note that the integrating factor for this differential equation is given as \( e^{-3} \int \frac{dy}{y} \), which simplifies to \( \frac{1}{y^3} \).
02

Define Functions M and N

The functions \( M \) and \( N \) are defined as follows: \( M = \frac{y^2 + xy^3}{y^3} = \frac{1}{y} + x \) and \( N = \frac{5y^2 - xy + y^3 \sin y}{y^3} = \frac{5}{y} - \frac{x}{y^2} + \sin y \).
03

Check for Exactness

To check for exactness, we find the partial derivatives \( M_y = - \frac{1}{y^2} \) and \( N_x = - \frac{1}{y^2} \). Since these partial derivatives are equal, the equation is exact.
04

Integrate \( f_x = M \)

Integrating \( f_x = \frac{1}{y} + x \) with respect to \( x \), we get \( f(x, y) = \frac{x}{y} + \frac{1}{2} x^2 + h(y) \), where \( h(y) \) is a function of \( y \).
05

Determine \( h(y) \) by Integrating \( h'(y) = N - \text{(terms in } f\text{ with } x \text{)} \)

From the expression for \( f \), compute \( h'(y) = 5/y + \sin y \). Integrating this with respect to \( y \), we obtain \( h(y) = 5 \ln |y| - \cos y \).
06

Write the General Solution

The general solution of the differential equation is obtained by combining \( f(x, y) \): \( \frac{x}{y} + \frac{1}{2} x^2 + 5 \ln |y| - \cos y = c \), where \( c \) is a constant of integration.

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Key Concepts

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

Integrating Factor
When dealing with differential equations, the integrating factor is an essential tool to simplify and solve them, especially when they are not easily separable. In this exercise, we identified the integrating factor by examining the form of the differential equation. The goal is to convert the differential equation into an exact equation, which is easier to solve.

You start by comparing the derived form \[\frac{(N_{x} - M_{y})}{M} = -\frac{3}{y}\]with a general expression to find the integrating factor. This results in the following integral:\[ e^{-3} \int \frac{dy}{y}\]which simplifies to \[\frac{1}{y^3}\].

Using this integrating factor transforms the equation so that it becomes exact, making further steps straightforward. This method is particularly useful when direct integration of the original equation seems challenging or impossible.
Partial Derivatives
Partial derivatives play a critical role in exact differential equations. They help check whether a differential equation is exact by balancing the mixed derivatives of the potential function. An exact differential equation has partial derivatives that satisfy a particular condition.

In this exercise, we defined functions \(M\) and \(N\) as:
  • \(M = \frac{y^2 + xy^3}{y^3} = \frac{1}{y} + x\)
  • \(N = \frac{5y^2 - xy + y^3 \sin y}{y^3} = \frac{5}{y} - \frac{x}{y^2} + \sin y\)
To check for exactness, we compare their partial derivatives:
  • \(M_y = - \frac{1}{y^2}\)
  • \(N_x = - \frac{1}{y^2}\)
Since \(M_y = N_x\), the equation is exact. This confirms the integrability condition and allows us to find the potential function, \(f(x, y)\). Always ensure the partial derivatives are equal to verify exactness, making the calculation of a general solution possible.
General Solution
Finding the general solution is the ultimate goal of solving differential equations, as it provides all possible solutions based on different initial conditions. The steps to achieve this involve both integration and pattern recognition.

Starting with the equation \(f_x = M\), you integrate with respect to \(x\) to find:\[f(x, y) = \frac{x}{y} + \frac{1}{2}x^2 + h(y)\]Identifying \(h(y)\) involves integrating the remaining part calculated from \(N - f_x\), which gives:
  • \(h'(y) = 5/y + \sin y\)
  • Integrating yields: \(h(y) = 5 \ln |y| - \cos y\)
Thus, the general solution becomes:\[\frac{x}{y} + \frac{1}{2}x^2 + 5 \ln |y| - \cos y = c\]where \(c\) is an arbitrary constant. This solution encompasses all particular solutions of the differential equation by varying \(c\), providing a complete set of solutions applicable to a range of boundary conditions.

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Most popular questions from this chapter

Separating variables, we have \\[\frac{d y}{y-y^{3}}=\frac{d y}{y(1-y)(1+y)}=\left(\frac{1}{y}+\frac{1 / 2}{1-y}-\frac{1 / 2}{1+y}\right) d y=d x\\]. Integrating, we get \\[\ln |y|-\frac{1}{2} \ln |1-y|-\frac{1}{2} \ln |1+y|=x+c\\]. When \(y>1,\) this becomes \\[\ln y-\frac{1}{2} \ln (y-1)-\frac{1}{2} \ln (y+1)=\ln \frac{y}{\sqrt{y^{2}-1}}=x+c.\\] Letting \(x=0\) and \(y=2\) we find \(c=\ln (2 / \sqrt{3}) .\) Solving for \(y\) we get \(y_{1}(x)=2 e^{x} / \sqrt{4 e^{2 x}-3},\) where \(x>\ln (\sqrt{3} / 2)\) When \(0\ln (\sqrt{3} / 2)\).

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 \(r(0)=0\).

Separating variables, we obtain \(d P / P=k \cos t d t,\) so $$\ln |P|=k \sin t+c \quad \text { and } \quad P=c_{1} e^{k \sin t}$$ If \(P(0)=P_{0},\) then \(c_{1}=P_{0}\) and \(P=P_{0} e^{k \sin t}\).

(a) From \(m d v / d t=m g-k v\) we obtain \(v=m g / k+c e^{-k t / m} .\) If \(v(0)=v_{0}\) then \(c=v_{0}-m g / k\) and the solution of the initial-value problem is $$v(t)=\frac{m g}{k}+\left(v_{0}-\frac{m g}{k}\right) e^{-k t / m}$$. (b) As \(t \rightarrow \infty\) the limiting velocity is \(m g / k\) (c) From \(d s / d t=v\) and \(s(0)=0\) we obtain $$s(t)=\frac{m g}{k} t-\frac{m}{k}\left(v_{0}-\frac{m g}{k}\right) e^{-k t / m}+\frac{m}{k}\left(v_{0}-\frac{m g}{k}\right)$$.

While the object is in the air its velocity is modelled by the linear differential equation \(m d v / d t=m g-k v .\) Using \(m=160, k=\frac{1}{4},\) and \(g=32,\) the differential equation becomes \(d v / d t+(1 / 640) v=32 .\) The integrating factor is \(e^{\int d t / 640}=e^{t / 640}\) and the solution of the differential equation is \(e^{t / 640} v=\int 32 e^{t / 640} d t=20,480 e^{t / 640}+c\) Using \(v(0)=0\) we see that \(c=-20,480\) and \(v(t)=20,480-20,480 e^{-t / 640} .\) Integrating we get \(s(t)=20,480 t+\) \(13,107,200 e^{-t / 640}+c .\) since \(s(0)=0, c=-13,107,200\) and \(s(t)=-13,107,200+20,480 t+13,107,200 e^{-t / 640}\) To find when the object hits the liquid we solve \(s(t)=500-75=425,\) obtaining \(t_{a}=5.16018 .\) The velocity at the time of impact with the liquid is \(v_{a}=v\left(t_{a}\right)=164.482 .\) When the object is in the liquid its velocity is modeled by the nonlinear differential equation \(m d v / d t=m g-k v^{2} .\) Using \(m=160, g=32,\) and \(k=0.1\) this becomes \(d v / d t=\left(51,200-v^{2}\right) / 1600 .\) Separating variables and integrating we have \\[ \frac{d v}{51,200-v^{2}}=\frac{d t}{1600} \quad \text { and } \quad \frac{\sqrt{2}}{640} \ln \left|\frac{v-160 \sqrt{2}}{v+160 \sqrt{2}}\right|=\frac{1}{1600} t+c \\] Solving \(v(0)=v_{a}=164.482\) we obtain \(c=-0.00407537 .\) Then, for \(v<160 \sqrt{2}=226.274\) \\[ \left|\frac{v-160 \sqrt{2}}{v+160 \sqrt{2}}\right|=e^{\sqrt{2} t / 5-1.8443} \quad \text { or } \quad-\frac{v-160 \sqrt{2}}{v+160 \sqrt{2}}=e^{\sqrt{2} t / 5-1.8443} \\] Solving for \(v\) we get \\[ v(t)=\frac{13964.6-2208.29 e^{\sqrt{2} t / 5}}{61.7153+9.75937 e^{\sqrt{2} t / 5}} \\] Integrating we find \\[ s(t)=226.275 t-1600 \ln \left(6.3237+e^{\sqrt{2} t / 5}\right)+c \\] Solving \(s(0)=0\) we see that \(c=3185.78,\) so \\[ s(t)=3185.78+226.275 t-1600 \ln \left(6.3237+e^{\sqrt{2} t / 5}\right) \\] To find when the object hits the bottom of the tank we solve \(s(t)=75,\) obtaining \(t_{b}=0.466273 .\) The time from when the object is dropped from the helicopter to when it hits the bottom of the tank is \(t_{a}+t_{b}=\) 5.62708 seconds.
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