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Chapter 17: Problem 1
Draw the direction field for \(y^{\prime}(t)=\sqrt{y(t)}\) and decide whether the equilibrium solution \(y(t)=0\) is stable.
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Which of the following possible solutions satisfies the initial condition and the differential equation. Possible Solution \(\quad\) Initial Condition \(\quad\) Differential Equation a. \(\quad y(t)=e^{2 t}+2 e^{t} \quad y(0)=2\) \(y^{\prime}(t)-y(t)=e^{2 t}\) b. \(\quad y(t)=e^{2 t}+2 e^{t} \quad y(0)=3 \quad y^{\prime}(t)-y(t)=e^{2 t}\) \(\begin{array}{lll}\text { c. } y(t)=\sqrt{t+1} & y(0)=1 & y(t) \times y^{\prime}(t)=1\end{array}\) d. \(y(t)=t^{3}\) \(y(1)=1 \quad y^{\prime}(t) / y(t)=3 t\) e. \(\quad y(t)=\frac{1}{t}\) f. \(y(t)=t^{3}\) \(y(1)=1 \quad y^{\prime}(t) / y(t)=3 t\) g. \(\quad y(t)=t e^{t}\) \(y(0)=0 \quad y^{\prime}(t)-y(t)=e^{t}\)
Show that the variables are not separable in the equation \(y^{\prime}(t)=t+y .\) That is, there are not two functions, \(g(t)\) and \(h(y),\) which for all \(t\) and \(y \quad t+y=g(t) \times h(y)\) A procedure is to assume two such functions, \(g(t)\) and \(h(y)\) exist and then show that the following equations are incompatible. \(\begin{array}{lllll}t=0 & y=0 & t=0 & y=1 & t=1 & y=0\end{array}\) \(g(0) \times h(0)=0+0=0 \quad g(0) \times h(1)=0+1=1 \quad g(1) \times h(0)=1+0=1\) Show that \(g(0) \times h(0)=0, g(0) \times h(1)=1\) and \(g(1) \times h(0)=1\) are incompatible.
Let \(M=10\) and \(r=0.1\) and plot the graphs of $$p(t)=\frac{M p_{0}}{p_{0}+\left(M-p_{0}\right) e^{-r t}}$$ for \(0 \leq t \leq 80\) and a. \(p_{0}=1\) $$\text { b. } p_{0}=12 \quad \text { c. } p_{0}=10$$
Show that each solution satisfies the initial condition and the differential equation. $$\begin{aligned} &\text { Solution } \quad \text { Initial Condition } \quad \text { Differential Equation }\\\ &\text { a. } \quad y(t)=e^{2 t}+e^{t} \quad y(0)=2 \quad y^{\prime}(t)-y(t)=e^{2 t} \end{aligned}$$ $$\text { b. } \quad y(t)=\frac{1}{3} e^{t}+\frac{2}{3} e^{-2 t}, \quad y(0)=1, \quad y^{\prime}(t)+2 y(t)=e^{t}$$ c. \(\quad y(t)=t e^{t} \quad y(0)=0 \quad y^{\prime}(t)-y(t)=e^{t}\) d. \(\quad y(t)=\frac{t^{2}}{3}+\frac{1}{t}, \quad y(1)=\frac{4}{3}, \quad t \times y^{\prime}(t)+y(t)=t^{2}\) e. \(\quad y(t)=\sqrt{t+1} \quad y(0)=1 \quad y(t) \times y^{\prime}(t)=\frac{1}{2}\) f. \(\quad y(t)=\sqrt{1+t^{2}}, \quad y(0)=1, \quad y(t) \times y^{\prime}(t)=t\) \(\begin{array}{lll}\text { g. } y(t)=\sqrt{4+t^{2}} & y(0)=2 & y(t) \times y^{\prime}(t)=t\end{array}\) \(\begin{array}{lll}\text { Solution } & \text { Initial } & \text { Differential Equation }\end{array}\) Condition h. \(y(t)=\frac{1}{t+1}, \quad y(0)=1, \quad y^{\prime}(t)+(y(t))^{2}=0\) i. \(\quad y(t)=0.5+0.5 e^{-0.2 \sin t} \quad y(0)=1, \quad y^{\prime}(t)+0.2(\cos t) y(t)=0.1 \cos t\) j. \(\quad y(t)=\tan t, \quad y(0)=0, \quad y^{\prime}(t)=1+(y(t))^{2}\) k. \(y(t)=3\) \(y(0)=3, \quad y^{\prime}(t)=(y(t)-1) \times(y(t)-3) \times(y(t)-5)\) l. \(y(t)=5\) \(y(0)=5, \quad y^{\prime}(t)=(y(t)-1) \times(y(t)-3) \times(y(t)-5)\)
Suppose immunity is not permanent in the SIR model, and recovered people become susceptible after six months. Modify the meaning of \(R\) and the SIR equations to account for this possibility.
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