91Ó°ÊÓ

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 \(0< x<\infty\).

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 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.

(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 \(I(15)=0.00098 I_{0}\).

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 \(y^{\prime}-\frac{1}{x} y=x \sin x\) 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 \(0< x<\infty\).

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 \(0< x<\infty\).