Chapter 2

Let \(M=y^{2}+y \sin x\) and \(N=2 x y-\cos x-1 /\left(1+y^{2}\right)\) so that \(M_{y}=2 y+\sin x=N_{x} .\) From \(f_{x}=y^{2}+y \sin x\) we obtain \(f=x y^{2}-y \cos x+h(y), h^{\prime}(y)=\frac{-1}{1+y^{2}},\) and \(h(y)=-\tan ^{-1} y .\) The solution is \(x y^{2}-y \cos x-\tan ^{-1} y=c\) If \(y(0)=1\) then \(c=-1-\pi / 4\) and a solution of the initial-value problem is \(x y^{2}-y \cos x-\tan ^{-1} y=-1-\frac{\pi}{4}\)

Separating variables and integrating we obtain \\[\frac{d x}{\sqrt{1-x^{2}}}-\frac{d y}{\sqrt{1-y^{2}}}=0 \quad \text { and } \quad \sin ^{-1} x-\sin ^{-1} y=c\\]. Setting \(x=0\) and \(y=\sqrt{3} / 2\) we obtain \(c=-\pi / 3 .\) Thus, an implicit solution of the initial-value problem is \(\sin ^{-1} x-\sin ^{-1} y=\pi / 3 .\) Solving for \(y\) and using an addition formula from trigonometry, we get \\[y=\sin \left(\sin ^{-1} x+\frac{\pi}{3}\right)=x \cos \frac{\pi}{3}+\sqrt{1-x^{2}} \sin \frac{\pi}{3}=\frac{x}{2}+\frac{\sqrt{3} \sqrt{1-x^{2}}}{2}\\].

For \(\frac{d i}{d t}+\frac{R}{L} i=\frac{E}{L}\) an integrating factor is \(e^{\int(R / L) d t}=e^{R t / L}\) so that \(\frac{d}{d t}\left[e^{R t / L} i\right]=\frac{E}{L} e^{R t / L}\) and \(i=\frac{E}{R}+c e^{-R t / L}\) for \(-\infty< t<\infty .\) If \(i(0)=i_{0}\) then \(c=i_{0}-E / R\) and \(i=\frac{E}{R}+\left(i_{0}-\frac{E}{R}\right) e^{-R t / L}\).

Equating \(M_{y}=3 y^{2}+4 k x y^{3}\) and \(N_{x}=3 y^{2}+40 x y^{3}\) we obtain \(k=10\)

Assume \(L d i / d t+R i=E(t), E(t)=E_{0} \sin \omega t,\) and \(i(0)=i_{0}\) so that $$i=\frac{E_{0} R}{L^{2} \omega^{2}+R^{2}} \sin \omega t-\frac{E_{0} L \omega}{L^{2} \omega^{2}+R^{2}} \cos \omega t+c e^{-R t / L}$$ since \(i(0)=i_{0}\) we obtain \(c=i_{0}+\frac{E_{0} L \omega}{L^{2} \omega^{2}+R^{2}}\)

From \(\frac{1}{1+(2 y)^{2}} d y=\frac{-x}{1+\left(x^{2}\right)^{2}} d x\) we obtain \(\frac{1}{2} \tan ^{-1} 2 y=-\frac{1}{2} \tan ^{-1} x^{2}+c \quad\) or \(\quad \tan ^{-1} 2 y+\tan ^{-1} x^{2}=c_{1}\). Using \(y(1)=0\) we find \(c_{1}=\pi / 4\). Thus, an implicit solution of the initial-value problem is \(\tan ^{-1} 2 y+\tan ^{-1} x^{2}=\pi / 4 .\) Solving for \(y\) and using a trigonometric identity we get $$\begin{aligned} 2 y &=\tan \left(\frac{\pi}{4}-\tan ^{-1} x^{2}\right) \\ y &=\frac{1}{2} \tan \left(\frac{\pi}{4}-\tan ^{-1} x^{2}\right) \\ &=\frac{1}{2} \frac{\tan \frac{\pi}{4}-\tan \left(\tan ^{-1} x^{2}\right)}{1+\tan \frac{\pi}{4} \tan \left(\tan ^{-1} x^{2}\right)} \\ &=\frac{1}{2} \frac{1-x^{2}}{1+x^{2}}. \end{aligned}$$

For \(\frac{d T}{d t}-k T=-T_{m} k\) an integrating factor is \(e^{\int(-k) d t}=e^{-k t}\) so that \(\frac{d}{d t}\left[e^{-k t} T\right]=-T_{m} k e^{-k t}\) and \(T=T_{m}+c e^{k t}\) for \(-\infty< t<\infty .\) If \(T(0)=T_{0}\) then \(c=T_{0}-T_{m}\) and \(T=T_{m}+\left(T_{0}-T_{m}\right) e^{k t}\).

Solving \(y e^{y}-9 y=y\left(e^{y}-9\right)=0\) we obtain the critical points 0 and \(\ln 9 .\) From the phase portrait we see that 0 is asymptotically stable (attractor) and \(\ln 9\) is unstable (repeller).

Assume \(R d q / d t+(1 / C) q=E(t), R=200, C=10^{-4},\) and \(E(t)=100\) so that \(q=1 / 100+c e^{-50 t} .\) If \(q(0)=0\) then \(c=-1 / 100\) and \(i=\frac{1}{2} e^{-50 t}\).

(a) The equilibrium solutions \(y(x)=2\) and \(y(x)=-2\) satisfy the initial conditions \(y(0)=2\) and \(y(0)=-2\) respectively. Setting \(x=\frac{1}{4}\) and \(y=1\) in \(y=2\left(1+c e^{4 x}\right) /\left(1-c e^{4 x}\right)\) we obtain $$1=2 \frac{1+c e}{1-c e}, \quad 1-c e=2+2 c e, \quad-1=3 c e, \quad \text { and } \quad c=-\frac{1}{3 e}$$. (b) Separating variables and integrating yields \\[ \begin{aligned} \frac{1}{4} \ln |y-2|-\frac{1}{4} \ln |y+2|+\ln c_{1} &=x \\ \ln |y-2|-\ln |y+2|+\ln c &=4 x \\ \ln \left|\frac{c(y-2)}{y+2}\right| &=4 x \\ c \frac{y-2}{y+2} &=e^{4 x}.\end{aligned}\\] Solving for \(y\) we get \(y=2\left(c+e^{4 x}\right) /\left(c-e^{4 x}\right) .\) The initial condition \(y(0)=-2\) implies \(2(c+1) /(c-1)=-2\) which yields \(c=0\) and \(y(x)=-2 .\) The initial condition \(y(0)=2\) does not correspond to a value of \(c,\) and it must simply be recognized that \(y(x)=2\) is a solution of the initial-value problem. Setting \(x=\frac{1}{4}\) and \(y=1\) in \(y=2\left(c+e^{4 x}\right) /\left(c-e^{4 x}\right)\) leads to \(c=-3 e .\) Thus, a solution of the initial-value problem is $$y=2 \frac{-3 e+e^{4 x}}{-3 e-e^{4 x}}=2 \frac{3-e^{4 x-1}}{3+e^{4 x-1}}.$$

Let \(u=x+y\) so that \(d u / d x=1+d y / d x .\) Then \(\frac{d u}{d x}-1=\cos u\) and \(\frac{1}{1+\cos u} d u=d x .\) Now $$\frac{1}{1+\cos u}=\frac{1-\cos u}{1-\cos ^{2} u}=\frac{1-\cos u}{\sin ^{2} u}=\csc ^{2} u-\csc u \cot u$$ so we have \(\int\left(\csc ^{2} u-\csc u \cot u\right) d u=\int d x\) and \(-\cot u+\csc u=x+c .\) Thus \(-\cot (x+y)+\csc (x+y)=x+c\) Setting \(x=0\) and \(y=\pi / 4\) we obtain \(c=\sqrt{2}-1 .\) The solution is $$\csc (x+y)-\cot (x+y)=x+\sqrt{2}-1.$$

For \(y^{\prime}+\frac{1}{x+1} y=\frac{\ln x}{x+1}\) an integrating factor is \(e^{\int[1 /(x+1)] d x}=x+1\) so that \(\frac{d}{d x}[(x+1) y]=\ln x\) and \(y=\frac{x}{x+1} \ln x-\frac{x}{x+1}+\frac{c}{x+1}\) for \(0< x<\infty .\) If \(y(1)=10\) then \(c=21\) and \(y=\frac{x}{x+1} \ln x-\frac{x}{x+1}+\frac{21}{x+1}\).

Separating variables, we have \\[\frac{d y}{y^{2}-y}=\frac{d x}{x} \quad \text { or } \quad \int \frac{d y}{y(y-1)}=\ln |x|+c.\\] Using partial fractions, we obtain $$\begin{aligned} \int\left(\frac{1}{y-1}-\frac{1}{y}\right) d y &=\ln |x|+c \\ \ln |y-1|-\ln |y| &=\ln |x|+c \\ \ln \left|\frac{y-1}{x y}\right| &=c \\ \frac{y-1}{x y} &=e^{c}=c_{1}.\end{aligned}$$ Solving for \(y\) we get \(y=1 /\left(1-c_{1} x\right) .\) We note by inspection that \(y=0\) is a singular solution of the differential equation. (a) Setting \(x=0\) and \(y=1\) we have \(1=1 /(1-0),\) which is true for all values of \(c_{1} .\) Thus, solutions passing through (0,1) are \(y=1 /\left(1-c_{1} x\right)\). (b) Setting \(x=0\) and \(y=0\) in \(y=1 /\left(1-c_{1} x\right)\) we get \(0=1 .\) Thus, the only solution passing through (0,0) is \(y=0\). (c) Setting \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\) we have \(\frac{1}{2}=1 /\left(1-\frac{1}{2} c_{1}\right),\) so \(c_{1}=-2\) and \(y=1 /(1+2 x)\). (d) \(\operatorname{Setting} x=2\) and \(y=\frac{1}{4}\) we have \(\frac{1}{4}=1 /\left(1-2 c_{1}\right),\) so \(c_{1}=-\frac{3}{2}\) and \(y=1 /\left(1+\frac{3}{2} x\right)=2 /(2+3 x)\).

For \(0 \leq t \leq 20\) the differential equation is \(20 d i / d t+2 i=120 .\) An integrating factor is \(e^{t / 10},\) so \((d / d t)\left[e^{t / 10} i\right]=\) \(6 e^{t / 10}\) and \(i=60+c_{1} e^{-t / 10} .\) If \(i(0)=0\) then \(c_{1}=-60\) and \(i=60-60 e^{-t / 10} .\) For \(t>20\) the differential equation is \(20 d i / d t+2 i=0\) and \(i=c_{2} e^{-t / 10} .\) At \(t=20\) we want \(c_{2} e^{-2}=60-60 e^{-2}\) so that \(c_{2}=60\left(e^{2}-1\right)\). Thus $$i(t)=\left\\{\begin{array}{ll} 60-60 e^{-t / 10}, & 0 \leq t \leq 20 \\ 60\left(e^{2}-1\right) e^{-t / 10}, & t>20 \end{array}\right.$$

For \(y^{\prime}+2 y=f(x)\) an integrating factor is \(e^{2 x}\) so that $$y e^{2 x}=\left\\{\begin{array}{ll} \frac{1}{2} e^{2 x}+c_{1}, & 0 \leq x \leq 3 \\ c_{2}, & x>3 \end{array}\right.$$ If \(y(0)=0\) then \(c_{1}=-1 / 2\) and for continuity we must have \(c_{2}=\frac{1}{2} e^{6}-\frac{1}{2}\) so that $$y=\left\\{\begin{array}{ll} \frac{1}{2}\left(1-e^{-2 x}\right), & 0 \leq x \leq 3 \\ \frac{1}{2}\left(e^{6}-1\right) e^{-2 x}, & x>3 \end{array}\right.$$

We note that \(\left(M_{y}-N_{x}\right) / N=1 / x,\) so an integrating factor is \(e^{\int d x / x}=x .\) Let \(M=2 x y^{2}+3 x^{2}\) and \(N=2 x^{2} y\) so that \(M_{y}=4 x y=N_{x} .\) From \(f_{x}=2 x y^{2}+3 x^{2}\) we obtain \(f=x^{2} y^{2}+x^{3}+h(y), h^{\prime}(y)=0,\) and \(h(y)=0 . \mathrm{A}\) solution of the differential equation is \(x^{2} y^{2}+x^{3}=c\)

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 $$F\left(\frac{y}{x}\right)=5\left(\frac{y}{x}\right)^{-1}-2\left(\frac{y}{x}\right).$$

Differentiating \(\ln \left(x^{2}+10\right)+\csc y=c\) we get \\[\begin{array}{c}\frac{2 x}{x^{2}+10}-\csc y \cot y \frac{d y}{d x}=0 \\\\\frac{2 x}{x^{2}+10}-\frac{1}{\sin y} \cdot \frac{\cos y}{\sin y} \frac{d y}{d x}=0\end{array}\\] or \\[2 x \sin ^{2} y d x-\left(x^{2}+10\right) \cos y d y=0\\] Writing the differential equation in the form \(\frac{d y}{d x}=\frac{2 x \sin ^{2} y}{\left(x^{2}+10\right) \cos y}\) we see that singular solutions occur when \(\sin ^{2} y=0,\) or \(y=k \pi,\) where \(k\) is an integer.

For \(y^{\prime}+y=f(x)\) an integrating factor is \(e^{x}\) so that $$y e^{x}=\left\\{\begin{array}{ll} e^{x}+c_{1}, & 0 \leq x \leq 1 \\ -e^{x}+c_{2}, & x>1 \end{array}\right.$$ If \(y(0)=1\) then \(c_{1}=0\) and for continuity we must have \(c_{2}=2 e\) so that $$y=\left\\{\begin{array}{ll} 1, & 0 \leq x \leq 1 \\ 2 e^{1-x}-1, & x>1 \end{array}\right.$$

(a) From \(m d v / d t=m g-k v\) we obtain \(v=m g / k+c e^{-k t / m} .\) If \(v(0)=v_{0}\) then \(c=v_{0}-m g / k\) and the solution of the initial-value problem is $$v(t)=\frac{m g}{k}+\left(v_{0}-\frac{m g}{k}\right) e^{-k t / m}$$. (b) As \(t \rightarrow \infty\) the limiting velocity is \(m g / k\) (c) From \(d s / d t=v\) and \(s(0)=0\) we obtain $$s(t)=\frac{m g}{k} t-\frac{m}{k}\left(v_{0}-\frac{m g}{k}\right) e^{-k t / m}+\frac{m}{k}\left(v_{0}-\frac{m g}{k}\right)$$.

(a) \(\mathrm{By}\) inspection \(y=x\) and \(y=-x\) are solutions of the differential equation and not members of the family \(y=x \sin \left(\ln x+c_{2}\right).\) (b) Letting \(x=5\) and \(y=0\) in \(\sin ^{-1}(y / x)=\ln x+c_{2}\) we get \(\sin ^{-1} 0=\ln 5+c\) or \(c=-\ln 5 .\) Then \(\sin ^{-1}(y / x)=\ln x-\ln 5=\ln (x / 5) .\) Because the range of the arcsine function is \([-\pi / 2, \pi / 2]\) we must have $$\begin{array}{c} -\frac{\pi}{2} \leq \ln \frac{x}{5} \leq \frac{\pi}{2} \\ e^{-\pi / 2} \leq \frac{x}{5} \leq e^{\pi / 2} \\ 5 e^{-\pi / 2} \leq x \leq 5 e^{\pi / 2}. \end{array}$$ The interval of definition of the solution is approximately [1.04, 24.05].

For \(y^{\prime}+2 x y=f(x)\) an integrating factor is \(e^{x^{2}}\) so that $$y e^{x^{2}}=\left\\{\begin{array}{ll} \frac{1}{2} e^{x^{2}}+c_{1}, & 0 \leq x \leq 1 \\ c_{2}, & x>1 \end{array}\right.$$ If \(y(0)=2\) then \(c_{1}=3 / 2\) and for continuity we must have \(c_{2}=\frac{1}{2} e+\frac{3}{2}\) so that $$y=\left\\{\begin{array}{ll} \frac{1}{2}+\frac{3}{2} e^{-x^{2}}, & 0 \leq x \leq 1 \\ \left(\frac{1}{2} e+\frac{3}{2}\right) e^{-x^{2}}, & x>1 \end{array}\right.$$

We note that \(\left(N_{x}-M_{y}\right) / M=2 / y,\) so an integrating factor is \(e^{\int 2 d y / y}=y^{2}\). Let \(M=6 x y^{3}\) and \(N=4 y^{3}+9 x^{2} y^{2}\) so that \(M_{y}=18 x y^{2}=N_{x} .\) From \(f_{x}=6 x y^{3}\) we obtain \(f=3 x^{2} y^{3}+h(y), h^{\prime}(y)=4 y^{3},\) and \(h(y)=y^{4} .\) A solution of the differential equation is \(3 x^{2} y^{3}+y^{4}=c\)

(a) Integrating \(d^{2} s / d t^{2}=-g\) we get \(v(t)=d s / d t=-g t+c .\) From \(v(0)=300\) we find \(c=300,\) and we are given \(g=32,\) so the velocity is \(v(t)=-32 t+300\). (b) Integrating again and using \(s(0)=0\) we get \(s(t)=-16 t^{2}+300 t .\) The maximum height is attained when \(v=0,\) that is, at \(t_{a}=9.375 .\) The maximum height will be \(s(9.375)=1406.25 \mathrm{ft}\).

As \(x \rightarrow-\infty, e^{6 x} \rightarrow 0\) and \(y \rightarrow 2 x+3 .\) Now write \(\left(1+c e^{6 x}\right) /\left(1-\alpha e^{6 x}\right)\) as \(\left(e^{-6 x}+c\right) /\left(e^{-6 x}-c\right) .\) Then, as \(x \rightarrow \infty, e^{-6 x} \rightarrow 0\) and \(y \rightarrow 2 x-3\).

Separating variables we obtain \(\frac{d y}{(y-1)^{2}}=d x .\) Then \(-\frac{1}{y-1}=x+c \quad\) and \(\quad y=\frac{x+c-1}{x+c}\). Setting \(x=0\) and \(y=1.01\) we obtain \(c=-100 .\) The solution is \\[y=\frac{x-101}{x-100}.\\]

For \\[ y^{\prime}+\frac{2 x}{1+x^{2}} y=\left\\{\begin{array}{ll} \frac{x}{1+x^{2}}, & 0 \leq x \leq 1 \\ \frac{-x}{1+x^{2}}, & x>1 \end{array}\right. \\] an integrating factor is \(1+x^{2}\) so that \\[ \left(1+x^{2}\right) y=\left\\{\begin{array}{ll} \frac{1}{2} x^{2}+c_{1}, & 0 \leq x \leq 1 \\ -\frac{1}{2} x^{2}+c_{2}, & x>1 \end{array}\right. \\] If \(y(0)=0\) then \(c_{1}=0\) and for continuity we must have \(c_{2}=1\) so that \\[ y=\left\\{\begin{array}{ll} \frac{1}{2}-\frac{1}{2\left(1+x^{2}\right)}, & 0 \leq x \leq 1 \\ \frac{3}{2\left(1+x^{2}\right)}-\frac{1}{2}, & x>1 \end{array}\right. \\]

We note that \(\left(M_{y}-N_{x}\right) / N=-\cot x,\) so an integrating factor is \(e^{-} \int \cot x d x=\csc x .\) Let \(M=\cos x \csc x=\cot x\) and \(N=(1+2 / y) \sin x \csc x=1+2 / y,\) so that \(M_{y}=0=N_{x} .\) From \(f_{x}=\cot x\) we obtain \(f=\ln (\sin x)+h(y)\) \(h^{\prime}(y)=1+2 / y,\) and \(h(y)=y+\ln y^{2} .\) A solution of the differential equation is \(\ln (\sin x)+y+\ln y^{2}=c\)

(a) The substitutions \(y=y_{1}+u\) and $$\frac{d y}{d x}=\frac{d y_{1}}{d x}+\frac{d u}{d x}$$ lead to $$\begin{aligned} \frac{d y_{1}}{d x}+\frac{d u}{d x} &=P+Q\left(y_{1}+u\right)+R\left(y_{1}+u\right)^{2} \\ &=P+Q y_{1}+R y_{1}^{2}+Q u+2 y_{1} R u+R u^{2} \end{aligned}$$ or $$\frac{d u}{d x}-\left(Q+2 y_{1} R\right) u=R u^{2}.$$ This is a Bernoulli equation with \(n=2\) which can be reduced to the linear equation $$\frac{d w}{d x}+\left(Q+2 y_{1} R\right) w=-R$$ by the substitution \(w=u^{-1}\) (b) Identify \(P(x)=-4 / x^{2}, Q(x)=-1 / x,\) and \(R(x)=1 .\) Then \(\frac{d w}{d x}+\left(-\frac{1}{x}+\frac{4}{x}\right) w=-1 .\) An integrating factor is \(x^{3}\) so that \(x^{3} w=-\frac{1}{4} x^{4}+c\) or \(u=\left[-\frac{1}{4} x+c x^{-3}\right]^{-1} .\) Thus, \(y=\frac{2}{x}+u\).

Separating variables we obtain \(\frac{d y}{(y-1)^{2}+0.01}=d x .\) Then \\[10 \tan ^{-1} 10(y-1)=x+c \text { and } y=1+\frac{1}{10} \tan \frac{x+c}{10}.\\] Setting \(x=0\) and \(y=1\) we obtain \(c=0 .\) The solution is \\[y=1+\frac{1}{10} \tan \frac{x}{10}.\\]

We note that \(\left(M_{y}-N_{x}\right) / N=3,\) so an integrating factor is \(e \int 3 d x=e^{3 x} .\) Let \(M=\left(10-6 y+e^{-3 x}\right) e^{3 x}=\) \(10 e^{3 x}-6 y e^{3 x}+1\) and \(N=-2 e^{3 x},\) so that \(M_{y}=-6 e^{3 x}=N_{x} .\) From \(f_{x}=10 e^{3 x}-6 y e^{3 x}+1\) we obtain \(f=\) \(\frac{10}{3} e^{3 x}-2 y e^{3 x}+x+h(y), h^{\prime}(y)=0,\) and \(h(y)=0 .\) A solution of the differential equation is \(\frac{10}{3} e^{3 x}-2 y e^{3 x}+x=c\)

Assuming that the air resistance is proportional to velocity and the positive direction is downward with \(s(0)=0\) the model for the velocity is \(m d v / d t=m g-k v .\) Using separation of variables to solve this differential equation, we obtain \(v(t)=m g / k+c e^{-k t / m} .\) Then, using \(v(0)=0,\) we get \(v(t)=(m g / k)\left(1-e^{-k t / m}\right)\) Letting \(k=0.5, m=(125+35) / 32=5,\) and \(g=32,\) we have \(v(t)=320\left(1-e^{-0.1 t}\right) .\) Integrating we find \(s(t)=320 t+3200 e^{-0.1 t}+c_{1} . \quad\) Solving \(s(0)=0\) for \(c_{1}\) we find \(c_{1}=-3200,\) therefore \(s(t)=\) \(320 t+3200 e^{-0.1 t}-3200 .\) At \(t=15,\) when the parachute opens, \(v(15)=248.598\) and \(s(15)=2314.02\) At this time the value of \(k\) changes to \(k=10\) and the new initial velocity is \(v_{0}=248.598 .\) With the parachute open, the skydiver's velocity is \(v_{p}(t)=m g / k+c_{2} e^{-k t / m},\) where \(t\) is reset to 0 when the parachute opens. Letting \(m=5, g=32,\) and \(k=10,\) this gives \(v_{p}(t)=16+c_{2} e^{-2 t} .\) From \(v(0)=248.598\) we find \(c_{2}=232.598\) so \(v_{p}(t)=16+232.598 e^{-2 t} .\) Integrating, we get \(s_{p}(t)=16 t-116.299 e^{-2 t}+c_{3} .\) Solving \(s_{p}(0)=0\) for \(c_{3}\) we find \(c_{3}=116.299,\) so \(s_{p}(t)=16 t-116.299 e^{-2 t}+116.299 .\) Twenty seconds after leaving the plane is five seconds after the parachute opens. The skydiver's velocity at this time is \(v_{p}(5)=16.0106 \mathrm{ft} / \mathrm{s}\) and she has fallen a total of \(s(15)+s_{p}(5)=2314.02+196.294=2510.31 \mathrm{ft} .\) Her terminal velocity is \(\lim _{t \rightarrow \infty} v_{p}(t)=16, \mathrm{s}\) she has very nearly reached her terminal velocity five seconds after the parachute opens. When the parachute opens, the distance to the ground is \(15,000-s(15)=15,000-2,314=12,686\) ft. Solving \(s_{p}(t)=12,686\) we get \(t=785.6 \mathrm{s}=13.1 \mathrm{min} .\) Thus, it will take her approximately 13.1 minutes to reach the ground after her parachute has opened and a total of \((785.6+15) / 60=13.34\) minutes after she exits the plane.

For \(y^{\prime}+e^{x} y=1\) an integrating factor is \(e^{e^{x}}\). Thus \\[ \frac{d}{d x}\left[e^{e^{x}} y\right]=e^{e^{x}} \quad \text { and } \quad e^{e^{x}} y=\int_{0}^{x} e^{e^{t}} d t+c \\] From \(y(0)=1\) we get \(c=e,\) so \(y=e^{-e^{x}} \int_{0}^{x} e^{e^{t}} d t+e^{1-e^{x}}\) When \(y^{\prime}+e^{x} y=0\) we can separate variables and integrate: \\[ \frac{d y}{y}=-e^{x} d x \text { and } \ln |y|=-e^{x}+c \\] Thus \(y=c_{1} e^{-e^{x}} .\) From \(y(0)=1\) we get \(c_{1}=e,\) so \(y=e^{1-e^{x}}\) When \(y^{\prime}+e^{x} y=e^{x}\) we can see by inspection that \(y=1\) is a solution.

We note that \(\left(N_{x}-M_{y}\right) / M=-3 / y,\) so an integrating factor is \(e^{-3} \int d y / y=1 / y^{3} .\) Let \(M=\left(y^{2}+x y^{3}\right) / y^{3}=\) \(1 / y+x\) and \(N=\left(5 y^{2}-x y+y^{3} \sin y\right) / y^{3}=5 / y-x / y^{2}+\sin y,\) so that \(M_{y}=-1 / y^{2}=N_{x} .\) From \(f_{x}=1 / y+x\) we obtain \(f=x / y+\frac{1}{2} x^{2}+h(y), h^{\prime}(y)=5 / y+\sin y,\) and \(h(y)=5 \ln |y|-\cos y .\) A solution of the differential equation is \(x / y+\frac{1}{2} x^{2}+5 \ln |y|-\cos y=c\)

Write the differential equation as $$\frac{d v}{d x}+\frac{1}{x} v=32 v^{-1},$$ and let \(u=v^{2}\) or \(v=u^{1 / 2} .\) Then $$\frac{d v}{d x}=\frac{1}{2} u^{-1 / 2} \frac{d u}{d x},$$ and substituting into the differential equation, we have $$\frac{1}{2} u^{-1 / 2} \frac{d u}{d x}+\frac{1}{x} u^{1 / 2}=32 u^{-1 / 2} \quad \text { or } \quad \frac{d u}{d x}+\frac{2}{x} u=64.$$ The latter differential equation is linear with integrating factor \(e^{\int(2 / x) d x}=x^{2},\) so $$\frac{d}{d x}\left[x^{2} u\right]=64 x^{2}$$ and $$x^{2} u=\frac{64}{3} x^{3}+c \quad \text { or } \quad v^{2}=\frac{64}{3} x+\frac{c}{x^{2}}.$$

Separating variables, we have \\[\frac{d y}{y-y^{3}}=\frac{d y}{y(1-y)(1+y)}=\left(\frac{1}{y}+\frac{1 / 2}{1-y}-\frac{1 / 2}{1+y}\right) d y=d x\\]. Integrating, we get \\[\ln |y|-\frac{1}{2} \ln |1-y|-\frac{1}{2} \ln |1+y|=x+c\\]. When \(y>1,\) this becomes \\[\ln y-\frac{1}{2} \ln (y-1)-\frac{1}{2} \ln (y+1)=\ln \frac{y}{\sqrt{y^{2}-1}}=x+c.\\] Letting \(x=0\) and \(y=2\) we find \(c=\ln (2 / \sqrt{3}) .\) Solving for \(y\) we get \(y_{1}(x)=2 e^{x} / \sqrt{4 e^{2 x}-3},\) where \(x>\ln (\sqrt{3} / 2)\) When \(0\ln (\sqrt{3} / 2)\).

An integrating factor for \(y^{\prime}-2 x y=1\) is \(e^{-x^{2}} .\) Thus \\[ \begin{aligned} \frac{d}{d x}\left[e^{-x^{2}} y\right] &=e^{-x^{2}} \\ e^{-x^{2}} y &=\int_{0}^{x} e^{-t^{2}} d t=\frac{\sqrt{\pi}}{2} \operatorname{erf}(x)+c \\ y &=\frac{\sqrt{\pi}}{2} e^{x^{2}} \operatorname{erf}(x)+c e^{x^{2}} \end{aligned} \\] From \(y(1)=(\sqrt{\pi} / 2) e \operatorname{erf}(1)+c e=1\) we get \(c=e^{-1}-\frac{\sqrt{\pi}}{2} \operatorname{erf}(1) .\) The solution of the initial-value problem is \\[ y=\frac{\sqrt{\pi}}{2} e^{x^{2}} \operatorname{erf}(x)+\left(e^{-1}-\frac{\sqrt{\pi}}{2} \operatorname{erf}(1)\right) e^{x^{2}} \\] \\[ =e^{x^{2}-1}+\frac{\sqrt{\pi}}{2} e^{x^{2}}(\operatorname{erf}(x)-\operatorname{erf}(1)) \\]

We note that \(\left(M_{y}-N_{x}\right) / N=2 x /\left(4+x^{2}\right),\) so an integrating factor is \(e^{-2 \int x d x /\left(4+x^{2}\right)}=1 /\left(4+x^{2}\right) .\) Let \(M=x /\left(4+x^{2}\right)\) and \(N=\left(x^{2} y+4 y\right) /\left(4+x^{2}\right)=y,\) so that \(M_{y}=0=N_{x} .\) From \(f_{x}=x\left(4+x^{2}\right)\) we obtain \(f=\frac{1}{2} \ln \left(4+x^{2}\right)+h(y), h^{\prime}(y)=y,\) and \(h(y)=\frac{1}{2} y^{2} .\) A solution of the differential equation is \(\frac{1}{2} \ln \left(4+x^{2}\right)+\frac{1}{2} y^{2}=c\)

Separating variables, we obtain \(d P / P=k \cos t d t,\) so $$\ln |P|=k \sin t+c \quad \text { and } \quad P=c_{1} e^{k \sin t}$$ If \(P(0)=P_{0},\) then \(c_{1}=P_{0}\) and \(P=P_{0} e^{k \sin t}\).

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 $$\begin{aligned} e^{a t} u &=\frac{b}{a} e^{a t}+c \\ e^{a t} P^{-1} &=\frac{b}{a} e^{a t}+c \\ P^{-1} &=\frac{b}{a}+c e^{-a t} \\ P &=\frac{1}{b / a+c e^{-a t}}=\frac{a}{b+c_{1} e^{-a t}}. \end{aligned}$$

(a) The second derivative of \(y\) is \\[\frac{d^{2} y}{d x^{2}}=-\frac{d y / d x}{(y-1)^{2}}=-\frac{1 /(y-3)}{(y-3)^{2}}=-\frac{1}{(y-3)^{3}}\\] The solution curve is concave down when \(d^{2} y / d x^{2}<0\) or \(y>3\), and concave up when \(d^{2} y / d x^{2}>0\) or \(y<3 .\) From the phase portrait we see that the solution curve is decreasing when \(y<3\) and increasing when \(y>3\). (b) Separating variables and integrating we obtain \\[ \begin{aligned} (y-3) d y &=d x \\ \frac{1}{2} y^{2}-3 y &=x+c \\ y^{2}-6 y+9 &=2 x+c_{1} \\ (y-3)^{2} &=2 x+c_{1} \\ y &=3 \pm \sqrt{2 x+c_{1}}.\end{aligned}\\] The initial condition dictates whether to use the plus or minus sign. When \(y_{1}(0)=4\) we have \(c_{1}=1\) and \(y_{1}(x)=3+\sqrt{2 x+1}\). When \(y_{2}(0)=2\) we have \(c_{1}=1\) and \(y_{2}(x)=3-\sqrt{2 x+1}.\) When \(y_{3}(1)=2\) we have \(c_{1}=-1\) and \(y_{3}(x)=3-\sqrt{2 x-1}.\) When \(y_{4}(-1)=4\) we have \(c_{1}=3\) and \(y_{4}(x)=3+\sqrt{2 x+3}\).

We note that \(\left(M_{y}-N_{x}\right) / N=-3 /(1+x),\) so an integrating factor is \(e^{-3 \int d x /(1+x)}=1 /(1+x)^{3}\). Let \(M=\) \(\left(x^{2}+y^{2}-5\right) /(1+x)^{3}\) and \(N=-(y+x y) /(1+x)^{3}=-y /(1+x)^{2},\) so that \(M_{y}=2 y /(1+x)^{3}=N_{x} .\) From \(f_{y}=-y /(1+x)^{2}\) we obtain \(f=-\frac{1}{2} y^{2} /(1+x)^{2}+h(x), h^{\prime}(x)=\left(x^{2}-5\right) /(1+x)^{3},\) and \(h(x)=2 /(1+x)^{2}+\) \(2 /(1+x)+\ln |1+x| .\) A solution of the differential equation is $$-\frac{y^{2}}{2(1+x)^{2}}+\frac{2}{(1+x)^{2}}+\frac{2}{(1+x)}+\ln |1+x|=c$$

(a) \(\operatorname{From} d P / d t=\left(k_{1}-k_{2}\right) P\) we obtain \(P=P_{0} e^{\left(k_{1}-k_{2}\right) t}\) where \(P_{0}=P(0)\). (b) If \(k_{1} >k_{2}\) then \(P \rightarrow \infty\) as \(t \rightarrow \infty\). If \(k_{1}=k_{2}\) then \(P=P_{0}\) for every \(t .\) If \(k_{1}< k_{2}\) then \(P \rightarrow 0\) as \(t \rightarrow \infty\).

(a) Separating variables we have \(2 y d y=(2 x+1) d x .\) Integrating gives \(y^{2}=x^{2}+x+c .\) When \(y(-2)=-1\) we find \(c=-1,\) so \(y^{2}=x^{2}+x-1\) and \(y=-\sqrt{x^{2}+x-1} .\) The negative square root is chosen because of the initial condition. (b) From the figure, the largest interval of definition appears to be approximately \((-\infty,-1.65)\). (c) Solving \(x^{2}+x-1=0\) we get \(x=-\frac{1}{2} \pm \frac{1}{2} \sqrt{5},\) so the largest interval of definition is \(\left(-\infty,-\frac{1}{2}-\frac{1}{2} \sqrt{5}\right)\) The right-hand endpoint of the interval is excluded because \(y=-\sqrt{x^{2}+x-1}\) is not differentiable at this point.

(a) All solutions of the form \(y=x^{5} e^{x}-x^{4} e^{x}+c x^{4}\) satisfy the initial condition. In this case, since \(4 / x\) is discontinuous at \(x=0,\) the hypotheses of Theorem 1.1 are not satisfied and the initial-value problem does not have a unique solution. (b) The differential equation has no solution satisfying \(y(0)=y_{0}, y_{0}>0\) (c) In this case, since \(x_{0}>0,\) Theorem 1.1 applies and the initial-value problem has a unique solution given by \(y=x^{5} e^{x}-x^{4} e^{x}+c x^{4}\) where \(c=y_{0} / x_{0}^{4}-x_{0} e^{x_{0}}+e^{x_{0}}\)

(a) Writing the differential equation in the form $$\frac{d v}{d t}=\frac{k}{m}\left(\frac{m g}{k}-v\right)$$ we see that a critical point is \(m g / k\). From the phase portrait we see that \(m g / k\) is an asymptotically stable critical point. Thus, \(\lim _{t \rightarrow \infty} v=m g / k\). (b) Writing the differential equation in the form $$\frac{d v}{d t}=\frac{k}{m}\left(\frac{m g}{k}-v^{2}\right)=\frac{k}{m}(\sqrt{\frac{m g}{k}}-v)(\sqrt{\frac{m g}{k}}+v)$$ we see that the only physically meaningful critical point is \(\sqrt{m g / k}\). From the phase portrait we see that \(\sqrt{m g / k}\) is an asymptotically stable critical point. Thus, \(\lim _{t \rightarrow \infty} v=\sqrt{m g / k}\).

(a) Solving \(k_{1}(M-A)-k_{2} A=0\) for \(A\) we find the equilibrium solution \(A=k_{1} M /\left(k_{1}+k_{2}\right) .\) From the phase portrait we see that \(\lim _{t \rightarrow \infty} A(t)=k_{1} M /\left(k_{1}+k_{2}\right)\) since \(k_{2}>0,\) the material will never be completely memorized and the larger \(k_{2}\) is, the less the amount of material will be memorized over time. (b) Write the differential equation in the form \(d A / d t+\left(k_{1}+k_{2}\right) A=k_{1} M\) Then an integrating factor is \(e^{\left(k_{1}+k_{2}\right) t},\) and $$\begin{aligned} \frac{d}{d t}\left[e^{\left(k_{1}+k_{2}\right) t} A\right] &=k_{1} M e^{\left(k_{1}+k_{2}\right) t} \\ e^{\left(k_{1}+k_{2}\right) t} A &=\frac{k_{1} M}{k_{1}+k_{2}} e^{\left(k_{1}+k_{2}\right) t}+c \\ A &=\frac{k_{1} M}{k_{1}+k_{2}}+c e^{-\left(k_{1}+k_{2}\right) t} \end{aligned}$$ \(\operatorname{Using} A(0)=0\) we find \(c=-\frac{k_{1} M}{k_{1}+k_{2}}\) and \(A=\frac{k_{1} M}{k_{1}+k_{2}}\left(1-e^{-\left(k_{1}+k_{2}\right) t}\right) .\) As \(t \rightarrow \infty\) \(A \rightarrow \frac{k_{1} M}{k_{1}+k_{2}}\).

(a) From the phase portrait we see that critical points are \(\alpha\) and \(\beta .\) Let \(X(0)=X_{0} .\) If \(X_{0} < \alpha\) we see that \(X \rightarrow \alpha\) as \(t \rightarrow \infty .\) If \(\alpha< X_{0} < \beta,\) we see that \(X \rightarrow \alpha\) as \(t \rightarrow \infty .\) If \(X_{0} > \beta,\) we see that \(X(t)\) increases in an unbounded manner, but more specific behavior of \(X(t)\) as \(t \rightarrow \infty\) is not known. (b) When \(\alpha=\beta\) the phase portrait is as shown. If \(X_{0}< \alpha,\) then \(X(t) \rightarrow \alpha\) as \(t \rightarrow \infty .\) If \(X_{0}> \alpha\) then \(X(t)\) increases in an unbounded manner. This could happen in a finite amount of time. That is, the phase portrait does not indicate that \(X\) becomes unbounded as \(t \rightarrow \infty\). (c) When \(k=1\) and \(\alpha=\beta\) the differential equation is \(d X / d t=(\alpha-X)^{2} .\) For \(X(t)=\alpha-1 /(t+c)\) we have \(d X / d t=1 /(t+c)^{2}\) and $$(\alpha-X)^{2}=\left[\alpha-\left(\alpha-\frac{1}{t+c}\right)\right]^{2}=\frac{1}{(t+c)^{2}}=\frac{d X}{d t}$$ For \(X(0)=\alpha / 2\) we obtain $$X(t)=\alpha-\frac{1}{t+2 / \alpha}$$ For \(X(0)=2 \alpha\) we obtain $$X(t)=\alpha-\frac{1}{t-1 / \alpha}$$ For \(X_{0}> \alpha, X(t)\) increases without bound up to \(t=1 / \alpha .\) For \(t> 1 / \alpha, X(t)\) increases but \(X \rightarrow \alpha\) as \(t \rightarrow \infty\).

(a) Solving \(r-k x=0\) for \(x\) we find the equilibrium solution \(x=r / k\). When \(x0\) and when \(x>r / k, d x / d t<0 .\) From the phase portrait we see that \(\lim _{t \rightarrow \infty} x(t)=r / k\). (b) From \(d x / d t=r-k x\) and \(x(0)=0\) we obtain \(x=r / k-(r / k) e^{-k t}\) so that \(x \rightarrow r / k\) as \(t \rightarrow \infty .\) If \(x(T)=r / 2 k\) then \(T=(\ln 2) / k\).

We want the general solution to be \(y=3 x-5+c e^{-x}\). (Rather than \(e^{-x}\), any function that approaches 0 as \(x \rightarrow \infty\) could be used.) Differentiating we get \\[ y^{\prime}=3-c e^{-x}=3-(y-3 x+5)=-y+3 x-2 \\] so the differential equation \(y^{\prime}+y=3 x-2\) has solutions asymptotic to the line \(y=3 x-5\).