Chapter 4

$5$

$y=2x^2+3x+2$

$y=\frac{1}{3}{x}^3-2x^2+3x-6e^x+14$

$v(t)=\mathrm{-9.8}t$

The equilibrium solutions are $y=\mathrm{-2}$ and $y=2.$ For this equation, $y=\mathrm{-2}$ is an unstable equilibrium solution, and $y=2$ is a semi-stable equilibrium solution.

$y=2+C{e}^{x^2+3x}$

$y=\frac{4+14e^{x^2+x}}{1-7e^{x^2+x}}$

Initial value problem:

$\frac{du}{dt}=2.4-\frac{2u}{25},u\left(0\right)=3$

$\text{Solution:}u\left(t\right)=30-27e^{\text{−}2t\text{/}25}$
$\text{Concentration: }30-27e^{\frac{\mathit{-}\mathit{2}\mathit{t}}{\mathit{25}}}$

1. Initial value problem
   $\frac{dT}{dt}=k\left(T-70\right),T\left(0\right)=450$
2. $T\left(t\right)=70+380e^{kt}$
3. Approximately $114$ minutes.

1. $\frac{dP}{dt}=0.04\left(1-\frac{P}{750}\right),P(0)=200$

2. 
   

3. $P\left(t\right)=\frac{3000e^{.04t}}{11+4e^{.04t}}$

4. After $12$ months, the population will be $P(12)\approx 278$ rabbits.

$y\prime +\frac{15}{x+3}y=\frac{10x-20}{x+3};p\left(x\right)=\frac{15}{x+3}$ and $q\left(x\right)=\frac{10x-20}{x+3}$

$y=\frac{x^3+x^2+C}{x-2}$

$y=-2x-\frac{5}{2}+\frac{1}{2}{e}^{2x}$

1. $\begin{array}{ccc}\hfill \frac{dv}{dt}& =\hfill & \text{−}v-9.8\hfill \\ \hfill v\left(0\right)& =\hfill & 0\hfill \end{array}$
2. $v\left(t\right)=9.8\left(e^{\text{−}t}-1\right)$
3. $\underset{t\to \infty }{\text{lim}}v\left(t\right)=\underset{t\to \infty }{\text{lim}}\left(9.8\left(e^{\text{−}t}-1\right)\right)=\mathrm{-9.8}\text{m/s}\approx -21.922\text{mph}$

Initial-value problem:

$8q^{\prime }+\frac{1}{0.02}q=20\text{sin}5t,q\left(0\right)=4$

$q\left(t\right)=\frac{10\text{sin}5t-8\text{cos}5t+172e^{\mathrm{-6.25}t}}{41}$

$1$

$3$

$1$

$1$

$y=4+\frac{3x^4}{4}$

$y=\frac{1}{2}{e}^{x^2}$

$y=2e^{\text{−}1\text{/}x}$

$u={\text{sin}}^{\mathrm{-1}}\left(e^{\mathrm{-1}+t}\right)$

$y=-\frac{\sqrt{x+1}}{\sqrt{1-x}}-1$

$y=C-x+x\text{ln}x-\text{ln}(\text{cos}x)$

$y=C+\frac{4^x}{\text{ln}(4)}$

$y=\frac{2}{3}\sqrt{t^2+16}\left(t^2+16\right)+C$

$x=\frac{2}{15}\sqrt{4+t}\left(3t^2+4t-32\right)+C$

$y=Cx$

$y=1-\frac{t^2}{2},y=-\frac{t^2}{2}-1$

$y=e^{\text{−}t},y=\text{−}{e}^{\text{−}t}$

$y=2\left(t^2+5\right),t=3\sqrt{5}$

$y=10e^{\mathrm{-2}t},t=-\frac{1}{2}\text{ln}\left(\frac{1}{10}\right)$

$y=\frac{1}{4}\left(41-e^{\mathrm{-4}t}\right),$ never

Solution changes from increasing to decreasing at $y(0)=0$

Solution changes from increasing to decreasing at $y(0)=0$

$v(t)=\mathrm{-32}t+a$

$0$ ft/s

$52.354$ meters

$x=50t-\frac{15}{{\pi }^2}\text{cos}(\pi t)+\frac{3}{{\pi }^2},2$ hours $1$ minute

$y=4e^{3t}$

$y=3-2t+t^2$

$y=\frac{1}{k}\left(e^{kt}-1\right)$ and $y=x$

$y=0$ is a stable equilibrium

$y=0$ is a stable equilibrium and $y=2$ is unstable

General solution is $y=e^t+C$.

General solution is $y=e^t(t-1)+C$.

E

A

B

A

C

$2.24,$ exact: $3$

$7.739364,$ exact: $5(e-1)$

$\mathrm{-0.2535}$ exact: $0$

$1.345,$ exact: $\frac{1}{\text{ln}(2)}$

$\mathrm{-4},$ exact: $\text{−}1\text{/}2$

$y\prime =2e^{t^2\text{/}2}$

$2$

$3.2756$

$2\sqrt{e}$

$4.0741e^{\mathrm{-10}}$

$y=e^t-1$

$y=1+C{e}^{\text{−}t}$

$y=Cx{e}^{\mathrm{-1}\text{/}x}$

$y=\frac{1}{C-x^2}$

$y=-\frac{2}{C+\text{ln}x}$

$y=C{e}^x\left(x+1\right)+1$

$y=\text{sin}\left(\text{ln}t+C\right)$

$y=\text{−}\text{ln}(e^{\text{−}x})$

$y=\frac{1}{\sqrt{2-e^{x^2}}}$

$y={\text{tanh}}^{\mathrm{-1}}\left(\frac{x^2}{2}\right)$

$x=\text{sin}\left(1-t+t\text{ln}t\right)$

$y=\text{ln}(\text{ln}(5))-\text{ln}(2-5^x)$

$y=C{e}^{\mathrm{-2}x}+\frac{1}{2}$

$y=\frac{1}{\sqrt{2}\sqrt{C-e^x}}$

$y=C{e}^{\text{−}x}{x}^x$

$y=\frac{r}{d}\left(1-e^{\text{−}dt}\right)$

$y(t)=10-9e^{\text{−}x\text{/}50}$

$134.3$ kilograms

$720$ seconds

$24$ hours $57$ minutes

$T(t)=20+50e^{\mathrm{-0.125}t}$

$T(t)=20+38.5e^{\mathrm{-0.125}t}$

$y=\left(c+\frac{b}{a}\right)e^{ax}-\frac{b}{a}$

$y(t)=cL+(I-cL)e^{\text{−}rt\text{/}L}$

$y=40\left(1-e^{\mathrm{-0.1}t}\right),40$ g/cm²

$P=0$ semi-stable

$P=\frac{10e^{10x}}{e^{10x}+4}$

$P\left(t\right)=\frac{10000e^{0.02t}}{150+50e^{0.02t}}$

$69$ hours $5$ minutes

$8$ years $11$ months

$P_1$ semi-stable

$P_2>0$ stable

$P_1=0$ is semi-stable

$y=\frac{\mathrm{-20}}{4\times {10}^{\mathrm{-6}}-0.002e^{0.01t}}$

$P\left(t\right)=\frac{850+500e^{0.009t}}{85+5e^{0.009t}}$

$13$ years months

$31.465$ days

September $2008$

$\frac{K+T}{2}$

$r=0.0405$

$\alpha =0.0081$

Logistic: $361,$ Threshold: $436,$ Gompertz: $309.$

Yes

Yes

$y\prime -x^{3}y=\text{sin}x$

$y\prime +\frac{\left(3x+2\right)}{x}y=\text{−}{e}^x$

$\frac{dy}{dt}-yx\left(x+1\right)=0$

$e^x$

$\text{−}\text{ln}\left(\text{cosh}x\right)$

$y=C{e}^{3x}-\frac{2}{3}$

$y=C{x}^3+6x^2$

$y=C{e}^{x^2\text{/}2}-3$

$y=C\text{tan}\left(\frac{x}{2}\right)-2x+4\text{tan}\left(\frac{x}{2}\right)\text{ln}\left(\text{sin}\left(\frac{x}{2}\right)\right)$

$y=C{x}^3-x^2$

$y=C{(x+2)}^2+\frac{1}{2}$

$y=\frac{C}{\sqrt{x}}+2\text{sin}(3t)$

$y=C{(x+1)}^3-x^2-2x-1$

$y=C{e}^{{\text{sinh}}^{\mathrm{-1}}x}-2$

$y=x+4e^x-1$

$y=-\frac{3x}{2}\left(x^2-1\right)$

$y=1-e^{{\text{tan}}^{\mathrm{-1}}x}$

$y=(x+2)\text{ln}\left(\frac{x+2}{2}\right)$

$y=2e^{2\sqrt{x}}-2x-2\sqrt{x}-1$

$v(t)=\frac{gm}{k}\left(1-e^{\text{−}kt\text{/}m}\right)$

$40.451$ seconds

$\sqrt{\frac{gm}{k}}$

$y=C{e}^x-a(x+1)$

$y=C{e}^{x^2\text{/}2}-a$

$y=\frac{e^{kt}-e^t}{k-1}$

F

T

$y\left(x\right)=\frac{2^x}{\text{ln}\left(2\right)}+x{\text{cos}}^{\mathrm{-1}}x-\sqrt{1-x^2}+C$

$y(x)=\text{ln}\left(C-\text{cos}x\right)$

$y(x)=e^{e^{C+x}}$

$y(x)=4+\frac{3}{2}{x}^2+2x-\text{sin}x$

$y(x)=-\frac{2}{1+3\left(x^2+2\text{sin}x\right)}$

$y(x)=\mathrm{-2}{x}^2-2x-\frac{1}{3}-\frac{2}{3}{e}^{3x}$

$y(x)=C{e}^{\text{−}x}+\text{ln}x$

Euler: $0.6939,$ exact solution: $y(x)=\frac{3^x-e^{\mathrm{-2}x}}{2+\text{ln}\left(3\right)}$

$\frac{40}{49}$ second

$x(t)=5000+\frac{245}{9}-\frac{49}{3}t-\frac{245}{9}{e}^{\text{−}5\text{/}3t},t=307.8$ seconds

$T(t)=200\left(1-e^{\text{−}t\text{/}1000}\right)$

$P(t)=\frac{1600000e^{0.02t}}{9840+160e^{0.02t}}$