Chapter 2

The linear equation \(d x / d t=-\lambda_{1} x\) can be solved by either separation of variables or by an integrating factor. Integrating both sides of \(d x / x=-\lambda_{1} d t\) we obtain \(\ln |x|=-\lambda_{1} t+c\) from which we get \(x=c_{1} e^{-\lambda_{1} t} .\) Using \(x(0)=x_{0}\) we find \(c_{1}=x_{0}\) so that \(x=x_{0} e^{-\lambda_{1} t} .\) Substituting this result into the second differential equation we have $$\frac{d y}{d t}+\lambda_{2} y=\lambda_{1} x_{0} e^{-\lambda_{1} t}$$ which is linear. An integrating factor is \(e^{\lambda_{2} t}\) so that $$\frac{d}{d t}\left[e^{\lambda_{2} t} y\right]=\lambda_{1} x_{0} e^{\left(\lambda_{2}-\lambda_{1}\right) t}+c_{2}$$ $$y=\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}} e^{\left(\lambda_{2}-\lambda_{1}\right) t} e^{-\lambda_{2} t}+c_{2} e^{-\lambda_{2} t}=\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{1} t}+c_{2} e^{-\lambda_{2} t}$$ Using \(y(0)=0\) we find \(c_{2}=-\lambda_{1} x_{0} /\left(\lambda_{2}-\lambda_{1}\right) .\) Thus $$y=\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}}\left(e^{-\lambda_{1} t}-e^{-\lambda_{2} t}\right)$$ Substituting this result into the third differential equation we have $$\frac{d z}{d t}=\frac{\lambda_{1} \lambda_{2} x_{0}}{\lambda_{2}-\lambda_{1}}\left(e^{-\lambda_{1} t}-e^{-\lambda_{2} t}\right)$$ Integrating we find $$z=-\frac{\lambda_{2} x_{0}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{1} t}+\frac{\lambda_{1} x_{0}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{2} t}+c_{3}$$ Using \(z(0)=0\) we find \(c_{3}=x_{0} .\) Thus $$z=x_{0}\left(1-\frac{\lambda_{2}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{1} t}+\frac{\lambda_{1}}{\lambda_{2}-\lambda_{1}} e^{-\lambda_{2} t}\right)$$.

(a) Solving \(N(1-0.0005 N)=0\) for \(N\) we find the equilibrium solutions \(N=0\) and \(N=2000\) When 0 From the phase portrait we see that \(\lim _{t \rightarrow \infty} N(t)=2000.\) (b) Separating variables and integrating we have \\[ \frac{d N}{N(1-0.0005 N)}=\left(\frac{1}{N}-\frac{1}{N-2000}\right) d N=d t \\] and \\[ \ln N-\ln (N-2000)=t+c \\] Solving for \(N\) we get \(N(t)=2000 e^{c+t} /\left(1+e^{c+t}\right)=2000 e^{c} e^{t} /\left(1+e^{c} e^{t}\right) .\) Using \(N(0)=1\) and solving for \(e^{c}\) we find \(e^{c}=1 / 1999\) and so \(N(t)=2000 e^{t} /\left(1999+e^{t}\right) .\) Then \(N(10)=1833.59,\) so 1834 companies are expected to adopt the new technology when \(t=10.\)

Let \(P=P(t)\) be the population at time \(t,\) and \(P_{0}\) the initial population. From \(d P / d t=k P\) we obtain \(P=P_{0} e^{k t}\) \(\operatorname{Using} P(5)=2 P_{0}\) we find \(k=\frac{1}{5} \ln 2\) and \(P=P_{0} e^{(\ln 2) t / 5} .\) Setting \(P(t)=3 P_{0}\) we have \(3=e^{(\ln 2) t / 5},\) so \(\ln 3=\frac{(\ln 2) t}{5} \quad\) and \(\quad t=\frac{5 \ln 3}{\ln 2} \approx 7.9\) years. Setting \(P(t)=4 P_{0}\) we have \(4=e^{(\ln 2) t / 5},\) so \(\ln 4=\frac{(\ln 2) t}{5} \quad\) and \(\quad t \approx 10\) years

We identify \(f(x, y)=2 x-3 y+1 .\) Then, for \(h=0.1\) $$y_{n+1}=y_{n}+0.1\left(2 x_{n}-3 y_{n}+1\right)=0.2 x_{n}+0.7 y_{n}+0.1$$ and $$\begin{aligned} &y(1.1) \approx y_{1}=0.2(1)+0.7(5)+0.1=3.8\\\ &y(1.2) \approx y_{2}=0.2(1.1)+0.7(3.8)+0.1=2.98 \end{aligned}$$ For \(h=0.05\) $$y_{n+1}=y_{n}+0.05\left(2 x_{n}-3 y_{n}+1\right)=0.1 x_{n}+0.85 y_{n}+0.1$$ and $$\begin{aligned} y(1.05) & \approx y_{1}=0.1(1)+0.85(5)+0.1=4.4 \\ y(1.1) & \approx y_{2}=0.1(1.05)+0.85(4.4)+0.1=3.895 \\ y(1.15) & \approx y_{3}=0.1(1.1)+0.85(3.895)+0.1=3.47075 \\ y(1.2) & \approx y_{4}=0.1(1.15)+0.85(3.47075)+0.1=3.11514 \end{aligned}$$

Letting \(y=u x\) we have $$\begin{aligned} (x-u x) d x+x(u d x+x d u) &=0 \\ d x+x d u &=0 \\ \frac{d x}{x}+d u &=0 \\ \ln |x|+u &=c \\ x \ln |x|+y &=c x. \end{aligned}$$

For \(y^{\prime}-5 y=0\) an integrating factor is \(e^{-\int 5 d x}=e^{-5 x}\) so that \(\frac{d}{d x}\left[e^{-5 x} y\right]=0\) and \(y=c e^{5 x}\) for \(-\infty < x < \infty\).

From Problem 1 the growth constant is \(k=\frac{1}{5} \ln 2 .\) Then \(P=P_{0} e^{(1 / 5)(\ln 2) t}\) and \(10,000=P_{0} e^{(3 / 5) \ln 2}\). Solving for \(P_{0}\) we get \(P_{0}=10,000 e^{-(3 / 5) \ln 2}=6,597.5 .\) Now $$P(10)=P_{0} e^{(1 / 5)(\ln 2)(10)}=6,597.5 e^{2 \ln 2}=4 P_{0}=26,390$$ The rate at which the population is growing is $$P^{\prime}(10)=k P(10)=\frac{1}{5}(\ln 2) 26,390=3658 \text { persons/year. }$$

For \(y^{\prime}+2 y=0\) an integrating factor is \(e^{\int 2 d x}=e^{2 x}\) so that \(\frac{d}{d x}\left[e^{2 x} y\right]=0\) and \(y=c e^{-2 x}\) for \(-\infty < x<\infty\). The transient term is \(c e^{-2 x}\).

From \(d y=(x+1)^{2} d x\) we obtain \(y=\frac{1}{3}(x+1)^{3}+c\).

From \(d P / d t=P\left(10^{-1}-10^{-7} P\right)\) and \(P(0)=5000\) we obtain \(P=500 /\left(0.0005+0.0995 e^{-0.1 t}\right)\) so that\ \(P \rightarrow 1,000,000\) as \(t \rightarrow \infty .\) If \(P(t)=500,000\) then \(t=52.9\) months.

Let \(P=P(t)\) be the population at time \(t .\) Then \(d P / d t=k P\) and \(P=c e^{k t} .\) From \(P(0)=c=500\) we see that \(P=500 e^{k t} .\) since \(15 \%\) of 500 is \(75,\) we have \(P(10)=500 e^{10 k}=575 .\) Solving for \(k,\) we get \(k=\frac{1}{10} \ln \frac{575}{500}=\frac{1}{10} \ln 1.15 .\) When \(t=30\), $$P(30)=500 e^{(1 / 10)(\ln 1.15) 30}=500 e^{3 \ln 1.15}=760 \text { years }$$ and $$P^{\prime}(30)=k P(30)=\frac{1}{10}(\ln 1.15) 760=10.62 \text { persons/year. }$$

Suppose that the series is described schematically by \(W \Longrightarrow-\lambda_{1} X \Longrightarrow-\lambda_{2} Y \Longrightarrow-\lambda_{3} Z\) where \(-\lambda_{1},-\lambda_{2},\) and \(-\lambda_{3}\) are the decay constants for \(W, X\) and \(Y,\) respectively, and \(Z\) is a stable element. Let \(w(t), x(t), y(t),\) and \(z(t)\) denote the amounts of substances \(W, X, Y,\) and \(Z,\) respectively. A model for the radioactive series is $$\begin{aligned}&\frac{d w}{d t}=-\lambda_{1} w\\\&\begin{array}{l} \frac{d x}{d t}=\lambda_{1} w-\lambda_{2} x \\\\\frac{d y}{d t}=\lambda_{2} x-\lambda_{3} y \end{array}\\\&\frac{d z}{d t}=\lambda_{3} y\end{aligned}$$

Separating variables and integrating, we have $$\frac{d y}{y}=2 x d x \text { and } \ln |y|=x^{2}+c$$ Thus \(y=c_{1} e^{x^{2}}\) and, using \(y(1)=1,\) we find \(c=e^{-1},\) so \(y=e^{x^{2}-1}\) is the solution of the initial-value problem.

(a) We have \(d P / d t=P(a-b P)\) with \(P(0)=3.929\) million. Using separation of variables we obtain \\[ \begin{aligned} P(t) &=\frac{3.929 a}{3.929 b+(a-3.929 b) e^{-a t}}=\frac{a / b}{1+(a / 3.929 b-1) e^{-a t}} \\ &=\frac{c}{1+(c / 3.929-1) e^{-a t}} \end{aligned} \\] where \(c=a / b .\) At \(t=60(1850)\) the population is 23.192 million, so \\[ 23.192=\frac{c}{1+(c / 3.929-1) e^{-60 a}} \\] or \(c=23.192+23.192(c / 3.929-1) e^{-60 a} .\) At \(t=120(1910)\) \\[ 91.972=\frac{c}{1+(c / 3.929-1) e^{-120 a}} \\] or \(c=91.972+91.972(c / 3.929-1)\left(e^{-60 a}\right)^{2}\). Combining the two equations for \(c\) we get \\[ \left(\frac{(c-23.192) / 23.192}{c / 3.929-1}\right)^{2}\left(\frac{c}{3.929}-1\right)=\frac{c-91.972}{91.972} \\] or \\[ 91.972(3.929)(c-23.192)^{2}=(23.192)^{2}(c-91.972)(c-3.929) \\] The solution of this quadratic equation is \(c=197.274 .\) This in turn gives \(a=0.0313 .\) Therefore, \\[ P(t)=\frac{197.274}{1+49.21 e^{-0.0313 t}} \\] $$\begin{array}{|c|rrrr|} \hline & {}{\underline{\phantom{xx}}} {\text { Census }} & {}{\underline{\phantom{xx}}} {\text { Predicted }} & {}{\underline{\phantom{xx}}} {} \\ \text { Year } & \text { Population } & \text { Population } & \text {} & \text { Error } \\ \hline 1790 & 3.929 & 3.929 & 0.000 & 0.00 \\ 1800 & 5.308 & 5.334 & -0.026 & -0.49 \\ 1810 & 7.240 & 7.222 & 0.018 & 0.24 \\ 1820 & 9.638 & 9.746 & -0.108 & -1.12 \\ 1830 & 12.866 & 13.090 & -0.224 & -1.74 \\ 1840 & 17.069 & 17.475 & -0.406 & -2.38 \\ 1850 & 23.192 & 23.143 & 0.049 & 0.21 \\ 1860 & 31.433 & 30.341 & 1.092 & 3.47 \\ 1870 & 38.558 & 39.272 & -0.714 & -1.85 \\ 1880 & 50.156 & 50.044 & 0.112 & 0.22 \\ 1890 & 62.948 & 62.600 & 0.348 & 0.55 \\ 1900 & 75.996 & 76.666 & -0.670 & -0.88 \\ 1910 & 91.972 & 91.739 & 0.233 & 0.25 \\ 1920 & 105.711 & 107.143 & -1.432 & -1.35 \\ 1930 & 122.775 & 122.140 & 0.635 & 0.52 \\ 1940 & 131.669 & 136.068 & -4.399 & -3.34 \\ 1950 & 150.697 & 148.445 & 2.252 & 1.49 \\ \hline \end{array}$$ The model predicts a population of 159.0 million for 1960 and 167.8 million for \(1970 .\) The census populations for these years were 179.3 and \(203.3,\) respectively. The percentage errors are 12.8 and \(21.2,\) respectively.

Let \(P=P(t)\) be bacteria population at time \(t\) and \(P_{0}\) the initial number. From \(d P / d t=k P\) we obtain \(P=P_{0} e^{k t} .\) Using \(P(3)=400\) and \(P(10)=2000\) we find \(400=P_{0} e^{3 k}\) or \(e^{k}=\left(400 / P_{0}\right)^{1 / 3} .\) From \(P(10)=2000\) we then have \(2000=P_{0} e^{10 k}=P_{0}\left(400 / P_{0}\right)^{10 / 3},\) so $$\frac{2000}{400^{10 / 3}}=P_{0}^{-7 / 3} \quad \text { and } \quad P_{0}=\left(\frac{2000}{400^{10 / 3}}\right)^{-3 / 7} \approx 201$$

Letting \(x=v y\) we have $$\begin{aligned} y(v d y+y d v)-2(v y+y) d y &=0 \\ y d v-(v+2) d y &=0 \\ \frac{d v}{v+2}-\frac{d y}{y} &=0 \\ \ln |v+2|-\ln |y| &=c \\ \ln \left|\frac{x}{y}+2\right|-\ln |y| &=c \\ x+2 y &=c_{1} y^{2}. \end{aligned}$$

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 \(x \sin y+y \cos x-\frac{1}{2} y^{2}=c\)

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

The system is $$\begin{aligned}&x_{1}^{\prime}=2 \cdot 3+\frac{1}{50} x_{2}-\frac{1}{50} x_{1} \cdot 4=-\frac{2}{25} x_{1}+\frac{1}{50} x_{2}+6\\\&x_{2}^{\prime}=\frac{1}{50} x_{1} \cdot 4-\frac{1}{50} x_{2}-\frac{1}{50} x_{2} \cdot 3=\frac{2}{25} x_{1}-\frac{2}{25} x_{2}\end{aligned}$$

(a) The differential equation is \(d P / d t=P(5-P)-4 .\) Solving \(P(5-P)-4=0\) for \(P\) we obtain equilibrium solutions \(P=1\) and \(P=4 .\) The phase portrait is shown on the right and solution curves are shown in part (b). We see that for \(P_{0}>4\) and \(1

Let \(A=A(t)\) be the amount of lead present at time \(t .\) From \(d A / d t=k A\) and \(A(0)=1\) we obtain \(A=e^{k t}\) \(\operatorname{Using} A(3.3)=1 / 2\) we find \(k=\frac{1}{3.3} \ln (1 / 2) .\) When \(90 \%\) of the lead has decayed, 0.1 grams will remain. Setting \(A(t)=0.1\) we have \(e^{t(1 / 3.3) \ln (1 / 2)}=0.1,\) so $$\frac{t}{3.3} \ln \frac{1}{2}=\ln 0.1 \quad \text { and } \quad t=\frac{3.3 \ln 0.1}{\ln (1 / 2)} \approx 10.96 \text { hours. }$$

From \(\frac{1}{y} d y=\frac{4}{x} d x\) we obtain \(\ln |y|=4 \ln |x|+c\) or \(y=c_{1} x^{4}\).

Solving \(P(5-P)-\frac{25}{4}=0\) for \(P\) we obtain the equilibrium solution \(P=\frac{5}{2} .\) For \(P \neq \frac{5}{2}, d P / d t<0 .\) Thus, if \(P_{0}<\frac{5}{2},\) the population becomes extinct (otherwise there would be another equilibrium solution.) Using separation of variables to solve the initial-value problem, we get \\[ P(t)=\left[4 P_{0}+\left(10 P_{0}-25\right) t\right] /\left[4+\left(4 P_{0}-10\right) t\right] \\] To find when the population becomes extinct for \(P_{0}<\frac{5}{2}\) we solve \(P(t)=0\) for \(t .\) We see that the time of extinction is \(t=4 P_{0} / 5\left(5-2 P_{0}\right).\)

Let \(A=A(t)\) be the amount present at time \(t .\) From \(d A / d t=k A\) and \(A(0)=100\) we obtain \(A=100 e^{k t} .\) Using \(A(6)=97\) we find \(k=\frac{1}{6} \ln 0.97 .\) Then \(A(24)=100 e^{(1 / 6)(\ln 0.97) 24}=100(0.97)^{4} \approx 88.5 \mathrm{mg}\).

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

Let \(M=4 x^{3}-3 y \sin 3 x-y / x^{2}\) and \(N=2 y-1 / x+\cos 3 x\) so that \(M_{y}=-3 \sin 3 x-1 / x^{2}\) and \(N_{x}=1 / x^{2}-3 \sin 3 x\) The equation is not exact.

From \(\frac{1}{y^{2}} d y=-2 x d x\) we obtain \(-\frac{1}{y}=-x^{2}+c\) or \(y=\frac{1}{x^{2}+c_{1}}\).

(a) A model is \\[\begin{array}{ll}\frac{d x_{1}}{d t}=3 \cdot \frac{x_{2}}{100-t}-2 \cdot \frac{x_{1}}{100+t}, & x_{1}(0)=100 \\ \frac{d x_{2}}{d t}=2 \cdot \frac{x_{1}}{100+t}-3 \cdot \frac{x_{2}}{100-t}, & x_{2}(0)=50 \end{array}\\] (b) since the system is closed, no salt enters or leaves the system and \(x_{1}(t)+x_{2}(t)=100+50=150\) for all time. Thus \(x_{1}=150-x_{2}\) and the second equation in part (a) becomes \\[\frac{d x_{2}}{d t}=\frac{2\left(150-x_{2}\right)}{100+t}-\frac{3 x_{2}}{100-t}=\frac{300}{100+t}-\frac{2 x_{2}}{100+t}-\frac{3 x_{2}}{100-t}\\] or \\[\frac{d x_{2}}{d t}+\left(\frac{2}{100+t}+\frac{3}{100-t}\right) x_{2}=\frac{300}{100+t}\\] Using integration by parts, we obtain \\[(100+t)^{2}(100-t)^{-3} x_{2}=300\left[\frac{1}{2}(100+t)(100-t)^{-2}-\frac{1}{2}(100-t)^{-1}+c\right]\\] Thus \\[\begin{aligned}x_{2} &=\frac{300}{(100+t)^{2}}\left[c(100-t)^{3}-\frac{1}{2}(100-t)^{2}+\frac{1}{2}(100+t)(100-t)\right] \\\ &=\frac{300}{(100+t)^{2}}\left[c(100-t)^{3}+t(100-t)\right]\end{aligned}\\] Using \(x_{2}(0)=50\) we find \(c=5 / 3000 .\) At \(t=30, x_{2}=\left(300 / 130^{2}\right)\left(70^{3} c+30 \cdot 70\right) \approx 47.4\) lbs.

Letting \(y=u x\) we have $$\begin{aligned} (u x-x) d x-(u x+x)(u d x+x d u) &=0 \\ \left(u^{2}+1\right) d x+x(u+1) d u &=0 \\ \frac{d x}{x}+\frac{u+1}{u^{2}+1} d u &=0 \\ \ln |x|+\frac{1}{2} \ln \left(u^{2}+1\right)+\tan ^{-1} u &=c \\ \ln x^{2}\left(\frac{y^{2}}{x^{2}}+1\right)+2 \tan ^{-1} \frac{y}{x} &=c_{1} \\\ \ln \left(x^{2}+y^{2}\right)+2 \tan ^{-1} \frac{y}{x} &=c_{1}. \end{aligned}$$

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

Let \(M=x^{2}-y^{2}\) and \(N=x^{2}-2 x y\) so that \(M_{y}=-2 y\) and \(N_{x}=2 x-2 y .\) The equation is not exact.

A model is $$\begin{aligned}&\frac{d x_{1}}{d t}=(4 \mathrm{gal} / \mathrm{min})(0 \mathrm{lb} / \mathrm{gal})-(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{200} x_{1} \mathrm{lb} / \mathrm{gal}\right)\\\&\begin{array}{l}\frac{d x_{2}}{d t}=(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{200} x_{1} \mathrm{lb} / \mathrm{gal}\right)-(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{150} x_{2} \mathrm{lb} / \mathrm{gal}\right) \\\\\frac{d x_{3}}{d t}=(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{150} x_{2} \mathrm{lb} / \mathrm{gal}\right)-(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{100} x_{3} \mathrm{lb} / \mathrm{gal}\right)\end{array}\end{aligned}$$ or $$\begin{aligned}\frac{d x_{1}}{d t} &=-\frac{1}{50} x_{1} \\\\\frac{d x_{2}}{d t} &=\frac{1}{50} x_{1}-\frac{2}{75} x_{2} \\\\\frac{d x_{3}}{d t} &=\frac{2}{75} x_{2}-\frac{1}{25} x_{3}\end{aligned}$$ Over a long period of time we would expect \(x_{1}, x_{2},\) and \(x_{3}\) to approach 0 because the entering pure water should flush the salt out of all three tanks.

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

(a) The solution of \(d A / d t=k A\) is \(A(t)=A_{0} e^{k t} .\) Letting \(A=\frac{1}{2} A_{0}\) and solving for \(t\) we obtain the half-life \(T=-(\ln 2) / k\). (b) since \(k=-(\ln 2) / T\) we have $$A(t)=A_{0} e^{-(\ln 2) t / T}=A_{0} 2^{-t / T}$$. (c) Writing \(\frac{1}{8} A_{0}=A_{0} 2^{-t / T}\) as \(2^{-3}=2^{-t / T}\) and solving for \(t\) we get \(t=3 T .\) Thus, an initial amount \(A_{0}\) will decay to \(\frac{1}{8} A_{0}\) in three half-lives.

Let \(X=X(t)\) be the amount of \(C\) at time \(t\) and \(d X / d t=k(120-2 X)(150-X) .\) If \(X(0)=0\) and \(X(5)=10\) then \\[ X(t)=\frac{150-150 e^{180 k t}}{1-2.5 e^{180 k t}} \\] where \(k=.0001259\) and \(X(20)=29.3\) grams. Now by L'Hôpital's rule, \(X \rightarrow 60\) as \(t \rightarrow \infty,\) so that the amount of \(A \rightarrow 0\) and the amount of \(B \rightarrow 30\) as \(t \rightarrow \infty.\)

$$h-0.1$$ $$h=0.05$$

Let \(I=I(t)\) be the intensity, \(t\) the thickness, and \(I(0)=I_{0} .\) If \(d I / d t=k I\) and \(I(3)=0.25 I_{0},\) then \(I=I_{0} e^{k t}\) \(k=\frac{1}{3} \ln 0.25,\) and \(I(15)=0.00098 I_{0}\).

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

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

For \(y^{\prime}+\frac{2}{x} y=\frac{3}{x}\) an integrating factor is \(e^{\int(2 / x) d x}=x^{2}\) so that \(\frac{d}{d x}\left[x^{2} y\right]=3 x\) and \(y=\frac{3}{2}+c x^{-2}\) for \(0< x<\infty\).

(a) The initial-value problem is \(d h / d t=-8 A_{h} \sqrt{h} / A_{w}, h(0)=H\) Separating variables and integrating we have \\[ \frac{d h}{\sqrt{h}}=-\frac{8 A_{h}}{A_{w}} d t \quad \text { and } \quad 2 \sqrt{h}=-\frac{8 A_{h}}{A_{w}} t+c \\] Using \(h(0)=H\) we find \(c=2 \sqrt{H},\) so the solution of the initial-value problem is \(\sqrt{h(t)}=\left(A_{w} \sqrt{H}-4 A_{h} t\right) / A_{w},\) where \(A_{w} \sqrt{H}-4 A_{h} t \geq 0 .\) Thus, \\[ h(t)=\left(A_{w} \sqrt{H}-4 A_{h} t\right)^{2} / A_{w}^{2} \quad \text { for } \quad 0 \leq t \leq A_{w} H / 4 A_{h} \\] (b) Identifying \(H=10, A_{w}=4 \pi,\) and \(A_{h}=\pi / 576\) we have \(h(t)=t^{2} / 331,776-(\sqrt{5 / 2} / 144) t+10 .\) Solving \(h(t)=0\) we see that the tank empties in \(576 \sqrt{10}\) seconds or 30.36 minutes.

Assume that \(A=A_{0} e^{k t}\) and \(k=-0.00012378\). If \(A(t)=0.145 A_{0}\) then \(t \approx 15,600\) years.

Letting \(y=u x\) we have $$\begin{aligned} \left(x^{3}-u^{3} x^{3}\right) d x+u^{2} x^{3}(u d x+x d u) &=0 \\ d x+u^{2} x d u &=0 \\ \frac{d x}{x}+u^{2} d u &=0 \\ \ln |x|+\frac{1}{3} u^{3} &=c \\ 3 x^{3} \ln |x|+y^{3} &=c_{1} x^{3}. \end{aligned}$$ Using \(y(-1)=1\) we find \(c_{1}=1 / 2 .\) The solution of the initial-value problem is \(2 x^{4}=y^{2}+x^{2}\).

For \(y^{\prime}+\frac{4}{x} y=x^{2}-1\) an integrating factor is \(e^{\int(4 / x) d x}=x^{4}\) so that \(\frac{d}{d x}\left[x^{4} y\right]=x^{6}-x^{4}\) and \(y=\frac{1}{7} x^{3}-\frac{1}{5} x+c x^{-4}\) for \(0< x <\infty\).

From \(2 y d y=-\frac{\sin 3 x}{\cos ^{3} 3 x} d x\) or \(2 y d y=-\tan 3 x \sec ^{2} 3 x d x\) we obtain \(y^{2}=-\frac{1}{6} \sec ^{2} 3 x+c\).

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

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

From \(\frac{e^{y}}{\left(e^{y}+1\right)^{2}} d y=\frac{-e^{x}}{\left(e^{x}+1\right)^{3}} d x\) we obtain \(-\left(e^{y}+1\right)^{-1}=\frac{1}{2}\left(e^{x}+1\right)^{-2}+c\).

Letting \(y=u x\) we have $$\begin{aligned} \left(x+u x e^{u}\right) d x-x e^{u}(u d x+x d u) &=0 \\ d x-x e^{u} d u &=0 \\ \frac{d x}{x}-e^{u} d u &=0 \\ \ln |x|-e^{u}=c. \end{aligned}$$ Using \(y(1)=0\) we find \(c=-1 .\) The solution of the initial-value problem is \(\ln |x|=e^{y / x}-1\).

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