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Chapter 2: Problem 4

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

Short Answer

Expert verified
The solution is \( x \sin y+y \cos x - \frac{1}{2}y^2 = c \).

Step by step solution

01

Understand the Functions

Given two functions, \( M = \sin y - y \sin x \) and \( N = \cos x + x \cos y - y \). The goal is to verify their relationship and use them to find the potential function \( f \).
02

Check Condition for Exact Differential Equation

We are given that \( M_y = \cos y - \sin x \) and \( N_x = \cos y - \sin x \), implying that \( M_y = N_x \). This confirms the equation is exact, and a potential function \( f(x, y) \) exists such that \( f_x = M \) and \( f_y = N \).
03

Integrate to Find Potential Function

Start by integrating \( f_x = M = \sin y - y \sin x \) with respect to \( x \). The result is \( f(x, y) = x \sin y + y \cos x + h(y) \), where \( h(y) \) is an arbitrary function of \( y \).
04

Differentiate with Respect to y

To find \( h(y) \), differentiate \( f(x, y) = x \sin y + y \cos x + h(y) \) with respect to \( y \) to get \( f_y = x \cos y + \cos x + h'(y) \).
05

Match with Given N

Given \( f_y = N = \cos x + x \cos y - y \), equate this to the result from Step 4: \( x \cos y + \cos x + h'(y) = \cos x + x \cos y - y \).
06

Solve for h(y)

By comparing, \( h'(y) = -y \). Integrate to find \( h(y) = -\frac{1}{2}y^2 + C \). Since constant \( C \) is usually absorbed in the arbitrary constant for the solution set, we set \( C=0 \).
07

Write the General Solution

The solution to the differential equation is \( f(x, y) = x \sin y + y \cos x - \frac{1}{2}y^2 = c \), where \( c \) is an arbitrary constant.

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Key Concepts

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

Potential Function
When dealing with exact differential equations, a potential function is a pivotal component. This function, often denoted as \( f(x, y) \), serves as an intermediary that consolidates the components of a differential equation. For our problem, the relationship between \( M = \sin y - y \sin x \) and \( N = \cos x + x \cos y - y \) is crucial. They must satisfy the equality \( M_y = N_x \) to confirm that a potential function exists. This equality, \( \cos y - \sin x = \cos y - \sin x \), proves that our equation is indeed exact. Consequently, the potential function \( f(x, y) \) can be constructed to satisfy \( f_x = M \) and \( f_y = N \). In this scenario, we have:
  • \( f_x = \sin y - y \sin x \)
  • \( f_y = \cos x + x \cos y - y \)
Finding the potential function sets up the pathway to solving the entire differential equation.
Integration
Integration plays a key role in uncovering the potential function from its partial derivatives. For this problem, we integrate the expression \( f_x = M = \sin y - y \sin x \) with respect to \( x \). This process results in the function \( f(x, y) = x \sin y + y \cos x + h(y) \). Here, \( h(y) \) is an arbitrary function that accounts for any terms solely dependent on \( y \), which ensures that the integration is complete.
  • Integrating \( \sin y - y \sin x \) with respect to \( x \) produces \( x \sin y + y \cos x \).
  • \( h(y) \) is an added function of \( y \) that is determined later by matching the original function \( N \).
Through integration, we start constructing the potential function piece by piece, getting closer to the full solution.
Differential Equation Verification
Verifying a differential equation involves ensuring the potential function accurately reflects both original functions \( M \) and \( N \). For verification, we differentiate our integrated potential function \( f(x, y) = x \sin y + y \cos x + h(y) \) with respect to \( y \). Achieving \( f_y = x \cos y + \cos x + h'(y) \), we compare this with \( N = \cos x + x \cos y - y \). The matching process reveals:
  • The terms \( x \cos y \) and \( \cos x \) match in both expressions.
  • This lets us conclude \( h'(y) = -y \), thus ensuring the functions are equivalent upon differentiation.
Differentiation and verification are critical; they ensure our potential function maintains coherence with the initial differential equation provided.
General Solution
The journey through potential functions and verification culminates in the general solution of the exact differential equation. In our exercise, solving for \( h(y) \) results in \( h(y) = -\frac{1}{2}y^2 + C \), where \( C \) serves as an arbitrary constant usually included in the constant of integration. Wrapping up these computations and integrations, we reach the elegant general solution: \( f(x, y) = x \sin y + y \cos x - \frac{1}{2}y^2 = c \). This final equation represents the family of solutions determined by any variation of \( c \). The general solution efficiently encompasses all potential behaviors of the system described by the differential equation. By understanding each step leading to it, students grasp how differential problems transform into complete mathematical representations, encapsulating infinitely many solutions.

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Most popular questions from this chapter

(a) \((i)\) Using Newton's second law of motion, \(F=m a=m d v / d t,\) the differential equation for the velocity \(v\) is $$m \frac{d v}{d t}=m g \sin \theta \quad \text { or } \quad \frac{d v}{d t}=g \sin \theta$$ where \(m g \sin \theta, 0< \theta< \pi / 2,\) is the component of the weight along the plane in the direction of motion. \((i i)\) The model now becomes $$m \frac{d v}{d t}=m g \sin \theta-\mu m g \cos \theta$$ where \(\mu m g \cos \theta\) is the component of the force of sliding friction (which acts perpendicular to the plane) along the plane. The negative sign indicates that this component of force is a retarding force which acts in the direction opposite to that of motion. \((i i i)\) If air resistance is taken to be proportional to the instantaneous velocity of the body, the model becomes $$m \frac{d v}{d t}=m g \sin \theta-\mu m g \cos \theta-k v$$ where \(k\) is a constant of proportionality. (b) \((i)\) With \(m=3\) slugs, the differential equation is $$3 \frac{d v}{d t}=(96) \cdot \frac{1}{2} \quad \text { or } \quad \frac{d v}{d t}=16$$ Integrating the last equation gives \(v(t)=16 t+c_{1} .\) since \(v(0)=0,\) we have \(c_{1}=0\) and so \(v(t)=16 t\). \((i i)\) With \(m=3\) slugs, the differential equation is $$3 \frac{d v}{d t}=(96) \cdot \frac{1}{2}-\frac{\sqrt{3}}{4} \cdot(96) \cdot \frac{\sqrt{3}}{2} \quad \text { or } \quad \frac{d v}{d t}=4$$ In this case \(v(t)=4 t\). (iii) When the retarding force due to air resistance is taken into account, the differential equation for velocity \(v\) becomes $$3 \frac{d v}{d t}=(96) \cdot \frac{1}{2}-\frac{\sqrt{3}}{4} \cdot(96) \cdot \frac{\sqrt{3}}{2}-\frac{1}{4} v \quad \text { or } \quad 3 \frac{d v}{d t}=12-\frac{1}{4} v$$ The last differential equation is linear and has solution \(v(t)=48+c_{1} e^{-t / 12} .\) since \(v(0)=0,\) we find \(c_{1}=-48,\) so \(v(t)=48-48 e^{-t / 12}\).

(a) The initial temperature of the bath is \(T_{m}(0)=60^{\circ},\) so in the short term the temperature of the chemical, which starts at \(80^{\circ},\) should decrease or cool. Over time, the temperature of the bath will increase toward \(100^{\circ}\) since \(e^{-0.1 t}\) decreases from 1 toward 0 as \(t\) increases from \(0 .\) Thus, in the long term, the temperature of the chemical should increase or warm toward \(100^{\circ}\). (b) Adapting the model for Newton's law of cooling, we have $$\frac{d T}{d t}=-0.1\left(T-100+40 e^{-0.1 t}\right), \quad T(0)=80$$. Writing the differential equation in the form $$\frac{d T}{d t}+0.1 T=10-4 e^{-0.1 t}$$ we see that it is linear with integrating factor \(e^{\int 0.1 d t}=e^{0.1 t}\) Thus $$\begin{aligned} \frac{d}{d t}\left[e^{0.1 t} T\right] &=10 e^{0.1 t}-4 \\ e^{0.1 t} T &=100 e^{0.1 t}-4 t+c \end{aligned}$$ and $$T(t)=100-4 t e^{-0.1 t}+c e^{-0.1 t}$$ Now \(T(0)=80\) so \(100+c=80, c=-20\) and $$T(t)=100-4 t e^{-0.1 t}-20 e^{-0.1 t}=100-(4 t+20) e^{-0.1 t}$$ The thinner curve verifies the prediction of cooling followed by warming toward \(100^{\circ} .\) The wider curve shows the temperature \(T_{m}\) of the liquid bath.

(a) The equilibrium solutions \(y(x)=2\) and \(y(x)=-2\) satisfy the initial conditions \(y(0)=2\) and \(y(0)=-2\) respectively. Setting \(x=\frac{1}{4}\) and \(y=1\) in \(y=2\left(1+c e^{4 x}\right) /\left(1-c e^{4 x}\right)\) we obtain $$1=2 \frac{1+c e}{1-c e}, \quad 1-c e=2+2 c e, \quad-1=3 c e, \quad \text { and } \quad c=-\frac{1}{3 e}$$. (b) Separating variables and integrating yields \\[ \begin{aligned} \frac{1}{4} \ln |y-2|-\frac{1}{4} \ln |y+2|+\ln c_{1} &=x \\ \ln |y-2|-\ln |y+2|+\ln c &=4 x \\ \ln \left|\frac{c(y-2)}{y+2}\right| &=4 x \\ c \frac{y-2}{y+2} &=e^{4 x}.\end{aligned}\\] Solving for \(y\) we get \(y=2\left(c+e^{4 x}\right) /\left(c-e^{4 x}\right) .\) The initial condition \(y(0)=-2\) implies \(2(c+1) /(c-1)=-2\) which yields \(c=0\) and \(y(x)=-2 .\) The initial condition \(y(0)=2\) does not correspond to a value of \(c,\) and it must simply be recognized that \(y(x)=2\) is a solution of the initial-value problem. Setting \(x=\frac{1}{4}\) and \(y=1\) in \(y=2\left(c+e^{4 x}\right) /\left(c-e^{4 x}\right)\) leads to \(c=-3 e .\) Thus, a solution of the initial-value problem is $$y=2 \frac{-3 e+e^{4 x}}{-3 e-e^{4 x}}=2 \frac{3-e^{4 x-1}}{3+e^{4 x-1}}.$$

(a) Separating variables and integrating, we have \\[\left(-2 y+y^{2}\right) d y=\left(x-x^{2}\right) d x\\] and \\[-y^{2}+\frac{1}{3} y^{3}=\frac{1}{2} x^{2}-\frac{1}{3} x^{3}+c\\]. Using a CAS we show some contours of \\[f(x, y)=2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}\\]. The plots shown on \([-7,7] \times[-5,5]\) correspond to \(c\) -values of -450,-300,-200,-120,-60,-20,-10,-8.1,-5 \(-0.8,20,60,\) and 120. (b) The value of \(c\) corresponding to \(y(0)=\frac{3}{2}\) is \(f\left(0, \frac{3}{2}\right)=-\frac{27}{4}\) The portion of the graph between the dots corresponds to the solution curve satisfying the intial condition. To determine the interval of definition we find \(d y / d x\) for \\[2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-\frac{27}{4}\\]. Using implicit differentiation we get \(y^{\prime}=\left(x-x^{2}\right) /\left(y^{2}-2 y\right)\) which is infinite when \(y=0\) and \(y=2 .\) Letting \(y=0\) in \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-\frac{27}{4}\) and using a CAS to solve for \(x\) we get \(x=-1.13232 .\) Similarly, letting \(y=2,\) we find \(x=1.71299 .\) The largest interval of definition is approximately (-1.13232,1.71299). (c) The value of \(c\) corresponding to \(y(0)=-2\) is \(f(0,-2)=-40\) The portion of the graph to the right of the dot corresponds to the solution curve satisfying the initial condition. To determine the interval of definition we find \(d y / d x\) for \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-40\). Using implicit differentiation we get \(y^{\prime}=\left(x-x^{2}\right) /\left(y^{2}-2 y\right)\) which is infinite when \(y=0\) and \(y=2 .\) Letting \(y=0\) in \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-40\) and using a CAS to solve for \(x\) we get \(x=-2.29551 .\) The largest interval of definition is approximately \((-2.29551, \infty)\).

For \(\frac{d i}{d t}+\frac{R}{L} i=\frac{E}{L}\) an integrating factor is \(e^{\int(R / L) d t}=e^{R t / L}\) so that \(\frac{d}{d t}\left[e^{R t / L} i\right]=\frac{E}{L} e^{R t / L}\) and \(i=\frac{E}{R}+c e^{-R t / L}\) for \(-\infty< t<\infty .\) If \(i(0)=i_{0}\) then \(c=i_{0}-E / R\) and \(i=\frac{E}{R}+\left(i_{0}-\frac{E}{R}\right) e^{-R t / L}\).
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