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Chapter 2: Problem 2
From \(d y=(x+1)^{2} d x\) we obtain \(y=\frac{1}{3}(x+1)^{3}+c\).
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Write the differential equation as \(d P / d t-a P=-b P^{2}\) and let \(u=P^{-1}\) or \(P=u^{-1}\). Then $$\frac{d p}{d t}=-u^{-2} \frac{d u}{d t},$$ and substituting into the differential equation, we have $$-u^{-2} \frac{d u}{d t}-a u^{-1}=-b u^{-2} \quad \text { or } \quad \frac{d u}{d t}+a u=b.$$ The latter differential equation is linear with integrating factor \(e^{\int a d t}=e^{a t},\) so $$\frac{d}{d t}\left[e^{a t} u\right]=b e^{a t}$$ and $$\begin{aligned} e^{a t} u &=\frac{b}{a} e^{a t}+c \\ e^{a t} P^{-1} &=\frac{b}{a} e^{a t}+c \\ P^{-1} &=\frac{b}{a}+c e^{-a t} \\ P &=\frac{1}{b / a+c e^{-a t}}=\frac{a}{b+c_{1} e^{-a t}}. \end{aligned}$$
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.
Rewrite \(\left(5 x^{2}-2 y^{2}\right) d x-x y d y=0\) as $$x y \frac{d y}{d x}=5 x^{2}-2 y^{2}$$ and divide by \(x y,\) so that $$\frac{d y}{d x}=5 \frac{x}{y}-2 \frac{y}{x}.$$ We then identify $$F\left(\frac{y}{x}\right)=5\left(\frac{y}{x}\right)^{-1}-2\left(\frac{y}{x}\right).$$
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.
The system is $$\begin{aligned}&x_{1}^{\prime}=2 \cdot 3+\frac{1}{50} x_{2}-\frac{1}{50} x_{1} \cdot 4=-\frac{2}{25} x_{1}+\frac{1}{50} x_{2}+6\\\&x_{2}^{\prime}=\frac{1}{50} x_{1} \cdot 4-\frac{1}{50} x_{2}-\frac{1}{50} x_{2} \cdot 3=\frac{2}{25} x_{1}-\frac{2}{25} x_{2}\end{aligned}$$
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