Chapter 2

When the height of the water is \(h,\) the radius of the top of the water is \(\frac{2}{5}(20-h)\) and \(A_{w}=4 \pi(20-h)^{2} / 25 .\) The differential equation is \\[ \frac{d h}{d t}=-c \frac{A_{h}}{A_{w}} \sqrt{2 g h}=-0.6 \frac{\pi(2 / 12)^{2}}{4 \pi(20-h)^{2} / 25} \sqrt{64 h}=-\frac{5}{6} \frac{\sqrt{h}}{(20-h)^{2}} \\] Separating variables and integrating we have \\[ \frac{(20-h)^{2}}{\sqrt{h}} d h=-\frac{5}{6} d t \quad \text { and } \quad 800 \sqrt{h}-\frac{80}{3} h^{3 / 2}+\frac{2}{5} h^{5 / 2}=-\frac{5}{6} t+c \\] Using \(h(0)=20\) we find \(c=2560 \sqrt{5} / 3,\) so an implicit solution of the initial-value problem is \\[ 800 \sqrt{h}-\frac{80}{3} h^{3 / 2}+\frac{2}{5} h^{5 / 2}=-\frac{5}{6} t+\frac{2560 \sqrt{5}}{3} \\] To find the time it takes the tank to empty we set \(h=0\) and solve for \(t .\) The tank empties in \(1024 \sqrt{5}\) seconds or 38.16 minutes. Thus, the tank empties more slowly when the base of the cone is on the bottom.

From \(\frac{y}{\left(1+y^{2}\right)^{1 / 2}} d y=\frac{x}{\left(1+x^{2}\right)^{1 / 2}} d x\) we obtain \(\left(1+y^{2}\right)^{1 / 2}=\left(1+x^{2}\right)^{1 / 2}+c\).

Assume that \(d T / d t=k(T-5)\) so that \(T=5+c e^{k t} .\) If \(T(1)=55^{\circ}\) and \(T(5)=30^{\circ}\) then \(k=-\frac{1}{4} \ln 2\) and \(c=59.4611\) so that \(T(0)=64.4611^{\circ}\).

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 \(r(0)=0\).

(a) After separating variables we obtain \\[ \begin{aligned} \frac{m d v}{m g-k v^{2}} &=d t \\ \frac{1}{g} \frac{d v}{1-(\sqrt{k} v / \sqrt{m g})^{2}} &=d t \\ \frac{\sqrt{m g}}{\sqrt{k} g} \frac{\sqrt{k / m g} d v}{1-(\sqrt{k} v / \sqrt{m g})^{2}} &=d t \\ \sqrt{\frac{m}{k g}} \tanh ^{-1} \frac{\sqrt{k} v}{\sqrt{m g}} &=t+c \\ \tanh ^{-1} \frac{\sqrt{k} v}{\sqrt{m g}} &=\sqrt{\frac{k g}{m}} t+c_{1} \end{aligned} \\] Thus the velocity at time \(t\) is \\[ v(t)=\sqrt{\frac{m g}{k}} \tanh \left(\sqrt{\frac{k g}{m}} t+c_{1}\right) \\] Setting \(t=0\) and \(v=v_{0}\) we find \(c_{1}=\tanh ^{-1}\left(\sqrt{k} v_{0} / \sqrt{m g}\right).\) (b) since \(\tanh t \rightarrow 1\) as \(t \rightarrow \infty,\) we have \(v \rightarrow \sqrt{m g / k}\) as \(t \rightarrow \infty\) (c) Integrating the expression for \(v(t)\) in part (a) we obtain an integral of the form \(\int d u / u:\) \\[ s(t)=\sqrt{\frac{m g}{k}} \int \tanh \left(\sqrt{\frac{k g}{m}} t+c_{1}\right) d t=\frac{m}{k} \ln \left[\cosh \left(\sqrt{\frac{k g}{m}} t+c_{1}\right)\right]+c_{2} \\] Setting \(t=0\) and \(s=0\) we find \(c_{2}=-(m / k) \ln \left(\cosh c_{1}\right),\) where \(c_{1}\) is given in part \((\mathrm{a}).\)

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.

From \(y^{\prime}+\frac{1}{x} y=\frac{1}{x} y^{-2}\) and \(w=y^{3}\) we obtain \(\frac{d w}{d x}+\frac{3}{x} w=\frac{3}{x} .\) An integrating factor is \(x^{3}\) so that \(x^{3} w=x^{3}+c\) or \(y^{3}=1+c x^{-3}\).

(a) If we know \(s(t)\) and \(i(t)\) then we can determine \(r(t)\) from \(s+i+r=n\) (b) In this case the system is \\[\begin{array}{l}\frac{d s}{d t}=-0.2 s i \\\\\frac{d i}{d t}=-0.7 i+0.2 s i\end{array}\\] We also note that when \(i(0)=i_{0}, s(0)=10-i_{0}\) since \(r(0)=0\) and \(i(t)+s(t)+r(t)=0\) for all values of \(t .\) Now \(k_{2} / k_{1}=0.7 / 0.2=3.5,\) so we consider initial conditions \(s(0)=2, i(0)=8 ; s(0)=3.4, i(0)=6.6\) \(s(0)=7, i(0)=3 ;\) and \(s(0)=9, i(0)=1\). We see that an initial susceptible population greater than \(k_{2} / k_{1}\) results in an epidemic in the sense that the number of infected persons increases to a maximum before decreasing to \(0 .\) On the other hand, when \(s(0)< k_{2} / k_{1},\) the number of infected persons decreases from the start and there is no epidemic.

From \(\frac{1}{Q-70} d Q=k d t\) we obtain \(\ln |Q-70|=k t+c\) or \(Q-70=c_{1} e^{k t}\).

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.

Let \(M=-2 y\) and \(N=5 y-2 x\) so that \(M_{y}=-2=N_{x} .\) From \(f_{x}=-2 y\) we obtain \(f=-2 x y+h(y), h^{\prime}(y)=5 y\) and \(h(y)=\frac{5}{2} y^{2} .\) A solution is \(-2 x y+\frac{5}{2} y^{2}=c\)

(a) Let \(\rho\) be the weight density of the water and \(V\) the volume of the object. Archimedes' principle states that the upward buoyant force has magnitude equal to the weight of the water displaced. Taking the positive direction to be down, the differential equation is \\[ m \frac{d v}{d t}=m g-k v^{2}-\rho V \\] (b) Using separation of variables we have \\[ \begin{aligned} \frac{m d v}{(m g-\rho V)-k v^{2}} &=d t \\ \frac{m}{\sqrt{k}} \frac{\sqrt{k} d v}{(\sqrt{m g-\rho V})^{2}-(\sqrt{k} v)^{2}} &=d t \\ \frac{m}{\sqrt{k}} \frac{1}{\sqrt{m g-\rho V}} \tanh ^{-1} \frac{\sqrt{k} v}{\sqrt{m g-\rho V}} &=t+c \end{aligned} \\] Thus \\[ v(t)=\sqrt{\frac{m g-\rho V}{k}} \tanh \left(\frac{\sqrt{k m g-k \rho V}}{m} t+c_{1}\right) \\] (c) since \(\tanh t \rightarrow 1\) as \(t \rightarrow \infty,\) the terminal velocity is \(\sqrt{(m g-\rho V) / k}.\)

Using separation of variables to solve \(d T / d t=k\left(T-T_{m}\right)\) we get \(T(t)=T_{m}+c e^{k t} .\) Using \(T(0)=70\) we find \(c=70-T_{m},\) so \(T(t)=T_{m}+\left(70-T_{m}\right) e^{k t} .\) Using the given observations, we obtain $$\begin{array}{c} T\left(\frac{1}{2}\right)=T_{m}+\left(70-T_{m}\right) e^{k / 2}=110 \\ T(1)=T_{m}+\left(70-T_{m}\right) e^{k}=145 \end{array}$$ Then, from the first equation, \(e^{k / 2}=\left(110-T_{m}\right) /\left(70-T_{m}\right)\) and $$\begin{aligned} e^{k}=\left(e^{k / 2}\right)^{2}=\left(\frac{110-T_{m}}{70-T_{m}}\right)^{2} &=\frac{145-T_{m}}{70-T_{m}} \\ \frac{\left(110-T_{m}\right)^{2}}{70-T_{m}} &=145-T_{m} \\ 12100-220 T_{m}+T_{m}^{2} &=10150-250 T_{m}+T_{m}^{2} \\ T_{m} &=390 \end{aligned}$$ The temperature in the oven is \(390^{\circ}\).

From \(\frac{1}{P-P^{2}} d P=\left(\frac{1}{P}+\frac{1}{1-P}\right) d P=d t\) we obtain \(\ln |P|-\ln |1-P|=t+c\) so that \(\ln \left|\frac{P}{1-P}\right|=t+c\) \(\frac{P}{1-P}=c_{1} e^{t} .\) Solving for \(P\) we have \(P=\frac{c_{1} e^{t}}{1+c_{1} e^{t}}\).

\text { For } y^{\prime}+(\tan x) y=\sec x \text { an integrating factor is } e^{\int \tan x d x}=\sec x \text { so that } \frac{d}{d x}[(\sec x) y]=\sec ^{2} x \text { and } \(y=\sin x+c \cos x\) for \(-\pi / 2< x<\pi / 2\).

(a) Writing the equation in the form \((x-\sqrt{x^{2}+y^{2}}) d x+y d y=0\) we identify \(M=x-\sqrt{x^{2}+y^{2}}\) and \(N=y\) since \(M\) and \(N\) are both homogeneous functions of degree 1 we use the substitution \(y=u x .\) It follows that \\[ \begin{aligned} (x-\sqrt{x^{2}+u^{2} x^{2}}) d x+u x(u d x+x d u) &=0 \\ x\left[1-\sqrt{1+u^{2}}+u^{2}\right] d x+x^{2} u d u &=0 \\ -\frac{u d u}{1+u^{2}-\sqrt{1+u^{2}}} &=\frac{d x}{x} \\ \frac{u d u}{\sqrt{1+u^{2}}(1-\sqrt{1+u^{2}})} &=\frac{d x}{x} \end{aligned} \\] Letting \(w=1-\sqrt{1+u^{2}}\) we have \(d w=-u d u / \sqrt{1+u^{2}}\) so that \\[ \begin{aligned} -\ln |1-\sqrt{1+u^{2}}| &=\ln |x|+c \\ \frac{1}{1-\sqrt{1+u^{2}}} &=c_{1} x \\ 1-\sqrt{1+u^{2}} &=-\frac{c_{2}}{x} \\ 1+\frac{c_{2}}{x} &=\sqrt{1+\frac{y^{2}}{x^{2}}} \\ 1+\frac{2 c_{2}}{x}+\frac{c_{2}^{2}}{x^{2}} &=1+\frac{y^{2}}{x^{2}} \end{aligned} \\] Solving for \(y^{2}\) we have \\[ y^{2}=2 c_{2} x+c_{2}^{2}=4\left(\frac{c_{2}}{2}\right)\left(x+\frac{c_{2}}{2}\right) \\] which is a family of parabolas symmetric with respect to the \(x\) -axis with vertex at \(\left(-c_{2} / 2,0\right)\) and focus at the origin. (b) Let \(u=x^{2}+y^{2}\) so that \\[ \frac{d u}{d x}=2 x+2 y \frac{d y}{d x} \\] Then \\[ y \frac{d y}{d x}=\frac{1}{2} \frac{d u}{d x}-x \\] and the differential equation can be written in the form \\[ \frac{1}{2} \frac{d u}{d x}-x=-x+\sqrt{u} \text { or } \frac{1}{2} \frac{d u}{d x}=\sqrt{u} \\] Separating variables and integrating gives $$\begin{aligned} \frac{d u}{2 \sqrt{u}} &=d x \\ \sqrt{u} &=x+c \\ u &=x^{2}+2 c x+c^{2} \\ x^{2}+y^{2} &=x^{2}+2 c x+c^{2} \\ y^{2} &=2 c x+c^{2} \end{aligned}$$

(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.

From \(\frac{1}{N} d N=\left(t e^{t+2}-1\right) d t\) we obtain \(\ln |N|=t e^{t+2}-e^{t+2}-t+c\) or \(N=c_{1} e^{t e^{t+2}-e^{t+2}-t}\).

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 \( x e^{x} w=-x e^{x}+e^{x}+c \text { or } y^{-1}=-1+\frac{1}{x}+\frac{c}{x} e^{-x}.\)

For \(y^{\prime}+(\cot x) y=\sec ^{2} x \csc x\) an integrating factor is \(e^{\int \cot x d x}=e^{\ln |\sin x|}=\sin x\) so that \(\frac{d}{d x}[(\sin x) y]=\sec ^{2} x\) and \(y=\sec x+c \csc x\) for \(0< x<\pi / 2\).

Let \(M=2 y \sin x \cos x-y+2 y^{2} e^{x y^{2}}\) and \(N=-x+\sin ^{2} x+4 x y e^{x y^{2}}\) so that $$M_{y}=2 \sin x \cos x-1+4 x y^{3} e^{x y^{2}}+4 y e^{x y^{2}}=N_{x}$$ From \(f_{x}=2 y \sin x \cos x-y+2 y^{2} e^{x y^{2}}\) we obtain \(f=y \sin ^{2} x-x y+2 e^{x y^{2}}+h(y), h^{\prime}(y)=0,\) and \(h(y)=0\) solution is \(y \sin ^{2} x-x y+2 e^{x y^{2}}=c\)

(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 \(d A / d t=4-A / 50\) we obtain \(A=200+c e^{-t / 50} .\) If \(A(0)=30\) then \(c=-170\) and \(A=200-170 e^{-t / 50}\).

For \(y^{\prime}+\frac{x+2}{x+1} y=\frac{2 x e^{-x}}{x+1}\) an integrating factor is \(e^{\int[(x+2) /(x+1)] d x}=(x+1) e^{x},\) so \(\frac{d}{d x}\left[(x+1) e^{x} y\right]=2 x\) and \(y=\frac{x^{2}}{x+1} e^{-x}+\frac{c}{x+1} e^{-x}\) for \(-1< x<\infty .\) The entire solution is transient.

Let \(M=4 t^{3} y-15 t^{2}-y\) and \(N=t^{4}+3 y^{2}-t\) so that \(M_{y}=4 t^{3}-1=N_{t} .\) From \(f_{t}=4 t^{3} y-15 t^{2}-y\) we obtain \(f=t^{4} y-5 t^{3}-t y+h(y), h^{\prime}(y)=3 y^{2},\) and \(h(y)=y^{3} .\) A solution is \(t^{4} y-5 t^{3}-t y+y^{3}=c\)

Writing the differential equation in the form \(d y / d x=y(1-y)(1+y)\) we see that critical points are located at \(y=-1, y=0,\) and \(y=1 .\) The phase portrait is shown at the right.

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 \(\frac{y+1}{y-1} d y=\frac{x+2}{x-3} d x\) or \(\left(1+\frac{2}{y-1}\right) d y=\left(1+\frac{5}{x-3}\right) d x\) we obtain \(y+2 \ln |y-1|=x+5 \ln |x-3|+c\) or \(\frac{(y-1)^{2}}{(x-3)^{5}}=c_{1} e^{x-y}\).

\text { For } y^{\prime}+\frac{4}{x+2} y=\frac{5}{(x+2)^{2}} \text { an integrating factor is } e^{\int[4 /(x+2)] d x}=(x+2)^{4} \text { so that } \frac{d}{d x}\left[(x+2)^{4} y\right]=5(x+2)^{2} and \(y=\frac{5}{3}(x+2)^{-1}+c(x+2)^{-4}\) for \(-2< x<\infty .\) The entire solution is transient.

From \(d A / d t=10-A / 100\) we obtain \(A=1000+c e^{-t / 100} .\) If \(A(0)=0\) then \(c=-1000\) and \(A(t)=\) \(1000-1000 e^{-t / 100}\).

From \(y^{\prime}-\frac{2}{x} y=\frac{3}{x^{2}} y^{4}\) and \(w=y^{-3}\) we obtain \(\frac{d w}{d x}+\frac{6}{x} w=-\frac{9}{x^{2}} .\) An integrating factor is \(x^{6}\) so that \(x^{6} w=-\frac{9}{5} x^{5}+c\) or \(y^{-3}=-\frac{9}{5} x^{-1}+c x^{-6} .\) If \(y(1)=\frac{1}{2}\) then \(c=\frac{49}{5}\) and \(y^{-3}=-\frac{9}{5} x^{-1}+\frac{49}{5} x^{-6}.\)

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

Let \(M=x^{2}+2 x y+y^{2}\) and \(N=2 x y+x^{2}-1\) so that \(M_{y}=2(x+y)=N_{x} .\) From \(f_{x}=x^{2}+2 x y+y^{2}\) we obtain \(f=\frac{1}{3} x^{3}+x^{2} y+x y^{2}+h(y), h^{\prime}(y)=-1,\) and \(h(y)=-y .\) The solution is \(\frac{1}{3} x^{3}+x^{2} y+x y^{2}-y=c .\) If \(y(1)=1\) then \(c=4 / 3\) and a solution of the initial-value problem is \(\frac{1}{3} x^{3}+x^{2} y+x y^{2}-y=\frac{4}{3}\)

From \(y^{\prime}+y=y^{-1 / 2}\) and \(w=y^{3 / 2}\) we obtain \(\frac{d w}{d x}+\frac{3}{2} w=\frac{3}{2} .\) An integrating factor is \(e^{3 x / 2}\) so that \(e^{3 x / 2} w=\) \(e^{3 x / 2}+c\) or \(y^{3 / 2}=1+c e^{-3 x / 2} .\) If \(y(0)=4\) then \(c=7\) and \(y^{3 / 2}=1+7 e^{-3 x / 2}\).

Let \(M=e^{x}+y\) and \(N=2+x+y e^{y}\) so that \(M_{y}=1=N_{x} . \quad\) From \(f_{x}=e^{x}+y\) we obtain \(f=e^{x}+x y+h(y), h^{\prime}(y)=2+y e^{y},\) and \(h(y)=2 y+y e^{y}-y .\) The solution is \(e^{x}+x y+2 y+y e^{y}-e^{y}=c .\) If \(y(0)=1\) then \(c=3\) and a solution of the initial-value problem is \(e^{x}+x y+2 y+y e^{y}-e^{y}=3\)

Let \(u=x+y+1\) so that \(d u / d x=1+d y / d x .\) Then \(\frac{d u}{d x}-1=u^{2}\) or \(\frac{1}{1+u^{2}} d u=d x .\) Thus \(\tan ^{-1} u=x+c\) or \(u=\tan (x+c),\) and \(x+y+1=\tan (x+c)\) or \(y=\tan (x+c)-x-1\).

From \(\frac{1}{x^{2}+1} d x=4 d t\) we obtain \(\tan ^{-1} x=4 t+c .\) Using \(x(\pi / 4)=1\) we find \(c=-3 \pi / 4 .\) The solution of the initial-value problem is \(\tan ^{-1} x=4 t-\frac{3 \pi}{4}\) or \(x=\tan \left(4 t-\frac{3 \pi}{4}\right)\).

Let \(M=4 y+2 t-5\) and \(N=6 y+4 t-1\) so that \(M_{y}=4=N_{t} .\) From \(f_{t}=4 y+2 t-5\) we obtain \(f=4 t y+t^{2}-5 t+h(y), h^{\prime}(y)=6 y-1,\) and \(h(y)=3 y^{2}-y .\) The solution is \(4 t y+t^{2}-5 t+3 y^{2}-y=c .\) If \(y(-1)=2\) then \(c=8\) and a solution of the initial-value problem is \(4 t y+t^{2}-5 t+3 y^{2}-y=8\)

(a) Solving \(v_{t}=\sqrt{m g / k}\) for \(k\) we obtain \(k=m g / v_{t}^{2} .\) The differential equation then becomes \\[ m \frac{d v}{d t}=m g-\frac{m g}{v_{t}^{2}} v^{2} \\] or \(\frac{d v}{d t}=g\left(1-\frac{1}{v_{t}^{2}} v^{2}\right)\) Separating variables and integrating gives \\[ v_{t} \tanh ^{-1} \frac{v}{v_{t}}=g t+c_{1} \\] The initial condition \(v(0)=0\) implies \(c_{1}=0,\) so \\[ v(t)=v_{t} \tanh \frac{g t}{v_{t}} \\] We find the distance by integrating: \\[ s(t)=\int v_{t} \tanh \frac{g t}{v_{t}} d t=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right)+c_{2} \\] The initial condition \(s(0)=0\) implies \(c_{2}=0,\) so \\[ s(t)=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right) \\] In 25 seconds she has fallen \(20,000-14,800=5,200\) feet. Using a CAS to solve \\[ 5200=\left(v_{t}^{2} / 32\right) \ln \left(\cosh \frac{32(25)}{v_{t}}\right) \\] for \(v_{t}\) gives \(v_{t} \approx 271.711 \mathrm{ft} / \mathrm{s} .\) Then \\[ s(t)=\frac{v_{t}^{2}}{g} \ln \left(\cosh \frac{g t}{v_{t}}\right)=2307.08 \ln (\cosh 0.117772 t) \\] (b) \(\operatorname{At} t=15, s(15)=2,542.94 \mathrm{ft}\) and \(v(15)=s^{\prime}(15)=256.287 \mathrm{ft} / \mathrm{sec}.\)

From \(\frac{1}{y^{2}-1} d y=\frac{1}{x^{2}-1} d x\) or \(\frac{1}{2}\left(\frac{1}{y-1}-\frac{1}{y+1}\right) d y=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right) d x\) we obtain \(\ln |y-1|-\ln |y+1|=\ln |x-1|-\ln |x+1|+\ln c\) or \(\frac{y-1}{y+1}=\frac{c(x-1)}{x+1} . \quad\) Using \(y(2)=2\) we find \(c=1 .\) A solution of the initial-value problem is \(\frac{y-1}{y+1}=\frac{x-1}{x+1}\) or \(y=x\).

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

Let \(M=t / 2 y^{4}\) and \(N=\left(3 y^{2}-t^{2}\right) / y^{5}\) so that \(M_{y}=-2 t / y^{5}=N_{t} .\) From \(f_{t}=t / 2 y^{4}\) we obtain \(f=\frac{t^{2}}{4 y^{4}}+h(y)\) \(h^{\prime}(y)=\frac{3}{y^{3}},\) and \(h(y)=-\frac{3}{2 y^{2}} .\) The solution is \(\frac{t^{2}}{4 y^{4}}-\frac{3}{2 y^{2}}=c .\) If \(y(1)=1\) then \(c=-5 / 4\) and a solution of the initial-value problem is \(\frac{t^{2}}{4 y^{4}}-\frac{3}{2 y^{2}}=-\frac{5}{4}\)

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.

Let \(u=x+y\) so that \(d u / d x=1+d y / d x .\) Then \(\frac{d u}{d x}-1=\tan ^{2} u\) or \(\cos ^{2} u d u=d x .\) Thus \(\frac{1}{2} u+\frac{1}{4} \sin 2 u=x+c\) or \(2 u+\sin 2 u=4 x+c_{1},\) and \(2(x+y)+\sin 2(x+y)=4 x+c_{1}\) or \(2 y+\sin 2(x+y)=2 x+c_{1}\).

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

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 \(y=\frac{1}{x} e^{x}+\frac{2-e}{x}\).

Let \(M=y^{2} \cos x-3 x^{2} y-2 x\) and \(N=2 y \sin x-x^{3}+\ln y\) so that \(M_{y}=2 y \cos x-3 x^{2}=N_{x} .\) From \(f_{x}=y^{2} \cos x-3 x^{2} y-2 x\) we obtain \(f=y^{2} \sin x-x^{3} y-x^{2}+h(y), h^{\prime}(y)=\ln y,\) and \(h(y)=y \ln y-y .\) The solution is \(y^{2} \sin x-x^{3} y-x^{2}+y \ln y-y=c .\) If \(y(0)=e\) then \(c=0\) and a solution of the initial-value problem is \(y^{2} \sin x-x^{3} y-x^{2}+y \ln y-y=0\)

Solving \(y^{2}\left(4-y^{2}\right)=y^{2}(2-y)(2+y)=0\) we obtain the critical points \(-2,0,\) and \(2 .\) From the phase portrait we see that 2 is asymptotically stable (attractor), 0 is semi-stable, and -2 is unstable (repeller).

(a) Initially the tank contains 300 gallons of solution. since brine is pumped in at a rate of 3 gal/min and the mixture is pumped out at a rate of 2 gal/min, the net change is an increase of 1 gal/min. Thus, in 100 minutes the tank will contain its capacity of 400 gallons. (b) The differential equation describing the amount of salt in the tank is \(A^{\prime}(t)=6-2 A /(300+t)\) with solution $$A(t)=600+2 t-\left(4.95 \times 10^{7}\right)(300+t)^{-2}, \quad 0 \leq t \leq 100$$ as noted in the discussion following Example 5 in the text. Thus, the amount of salt in the tank when it overflows is $$A(100)=800-\left(4.95 \times 10^{7}\right)(400)^{-2}=490.625 \mathrm{lbs}$$. (c) When the tank is overflowing the amount of salt in the tank is governed by the differential equation $$\begin{aligned} \frac{d A}{d t} &=(3 \mathrm{gal} / \mathrm{min})(2 \mathrm{lb} / \mathrm{gal})-\left(\frac{A}{400} \mathrm{lb} / \mathrm{gal}\right)(3 \mathrm{gal} / \mathrm{min}) \\ &=6-\frac{3 A}{400}, \quad A(100)=490.625 \end{aligned}$$ Solving the equation, we obtain \(A(t)=800+c e^{-3 t / 400} .\) The initial condition yields \(c=-654.947,\) so that $$A(t)=800-654.947 e^{-3 t / 400}$$. When \(t=150, A(150)=587.37\) lbs. (d) As \(t \rightarrow \infty,\) the amount of salt is 800 lbs, which is to be expected since \((400 \text { gal })(2 \mathrm{lb} / \mathrm{gal})=800 \mathrm{lbs}\) (e)

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 \(y=2 e^{-2 t}+\frac{1}{2}\).