# Quiz 9: Separable differential equations

Question

## Question 1

Find the solution to the differential equation
$\frac{dy}{dx}=2y\phantom{\rule{0.3em}{0ex}}.$
 a) $y={x}^{2}+C$ b) $y={e}^{\frac{1}{2}x}$ c) $y=A{e}^{2x}$ d) $y=\frac{1}{2}A{e}^{x}$

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
$\begin{array}{llll}\hfill \frac{dy}{dx}& =2y,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \int \frac{dy}{y}& =\int 2dx,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill ln|y|& =2x+C,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill |y|& ={e}^{2x+C}=B{e}^{2x}\phantom{\rule{1em}{0ex}}where\phantom{\rule{1em}{0ex}}B={e}^{C},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =A{e}^{2x}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Not correct. Choice (d) is false.

## Question 2

Find the solution to the differential equation
$\frac{dP}{dt}=3P\phantom{\rule{1em}{0ex}}where\phantom{\rule{1em}{0ex}}P=4\phantom{\rule{1em}{0ex}}when\phantom{\rule{1em}{0ex}}t=0\phantom{\rule{0.3em}{0ex}}.$
 a) $P=A{e}^{3t}$ b) $P=3A{e}^{t}$ c) $P=12A{e}^{t}$ d) $P=4{e}^{3t}$

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.

## Question 3

Find the solution to the differential equation
$\frac{dy}{dx}=xy\phantom{\rule{0.3em}{0ex}}.$
 a) $y=\frac{1}{2}{x}^{2}+C$ b) $y=B{e}^{\frac{1}{2}{x}^{2}}$ c) $y={e}^{\frac{1}{2}{x}^{2}}$ d) $y=\frac{1}{2}B{e}^{{x}^{2}}$

Not correct. Choice (a) is false.
$\begin{array}{llll}\hfill \frac{dy}{dx}& =xy,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \int \frac{dy}{y}& =\int x\phantom{\rule{0.3em}{0ex}}dx,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill ln|y|& =\frac{1}{2}{x}^{2}+C,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill |y|& ={e}^{\frac{1}{2}x+C}=A{e}^{\frac{1}{2}x}\phantom{\rule{1em}{0ex}}where\phantom{\rule{1em}{0ex}}A={e}^{C},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =B{e}^{\frac{1}{2}x}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

## Question 4

Find the solution to the differential equation
$\frac{dP}{dt}=4P+5\phantom{\rule{0.3em}{0ex}}.$
 a) $P=A{e}^{4t}+5t$ b) $P=A{e}^{4t}-1.25$ c) $P=5{e}^{4t}+5t$ d) $P=A{e}^{4t}+1.25t$

Not correct. Choice (a) is false.
$\begin{array}{llll}\hfill \frac{dP}{dt}& =4P+5=4\left(P+1.25\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \int \frac{dP}{\left(P+1.25\right)}& =\int 4\phantom{\rule{0.3em}{0ex}}dt,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill ln|P+1.25|& =4t+C,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill |P+1.25|& ={e}^{4t+C}=B{e}^{4t}\phantom{\rule{1em}{0ex}}where\phantom{\rule{1em}{0ex}}B={e}^{C},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill P+1.25& =A{e}^{4t},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill P& =A{e}^{4t}-1.25.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

## Question 5

Find the solution to the differential equation
$\frac{dP}{dt}=2P-2Pt\phantom{\rule{1em}{0ex}}where\phantom{\rule{1em}{0ex}}P=5\phantom{\rule{1em}{0ex}}when\phantom{\rule{1em}{0ex}}t=0\phantom{\rule{0.3em}{0ex}}.$
 a) $P=5{e}^{2-2t}$ b) $P=5{e}^{2t-{t}^{2}}$ c) $P=5{e}^{2-2{t}^{2}}$ d) $P={P}^{2}-{P}^{2}t$

Not correct. Choice (a) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

## Question 6

Find the solution to the differential equation
$\frac{dN}{dt}-\frac{N}{k}=0\phantom{\rule{1em}{0ex}}where\phantom{\rule{1em}{0ex}}k\phantom{\rule{1em}{0ex}}is\phantom{\rule{1em}{0ex}}a\phantom{\rule{1em}{0ex}}constant.$
 a) $N=B{e}^{\frac{t}{k}}$ where $B$ is a constant. b) $N=B{e}^{-kt}$ where $B$ is a constant. c) $N=B{e}^{-\frac{t}{k}}$ where $B$ is a constant. d) $N=B{e}^{kt}$ where $B$ is a constant.

$\begin{array}{llll}\hfill \frac{dN}{dt}& =\frac{N}{k},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \int \frac{dN}{N}& =\int \frac{1}{k},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill ln|N|& =\frac{t}{k}+C,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill |N|& =A{e}^{\frac{t}{k}}\phantom{\rule{1em}{0ex}}where\phantom{\rule{1em}{0ex}}A={e}^{C},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill N& =B{e}^{\frac{t}{k}}\phantom{\rule{1em}{0ex}}where\phantom{\rule{1em}{0ex}}B\phantom{\rule{1em}{0ex}}is\phantom{\rule{1em}{0ex}}a\phantom{\rule{1em}{0ex}}constant.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

## Question 7

Find the family of curves which satisfy the differential equation $\frac{dy}{dx}=-\frac{x}{y}\phantom{\rule{0.3em}{0ex}}.$
 a) ${x}^{2}+{y}^{2}={r}^{2}$ b) $y=\sqrt{{x}^{2}+C}$ c) $y=-\frac{1}{2}{x}^{2}+C$ d) This differential equation cannot be solved using the techniques learned in this course.

Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

## Question 8

Find the family of curves which satisfy
$\frac{dy}{dx}=3{x}^{2}\left(y-1\right).$
 a) $y=B{e}^{{x}^{3}}-1$ where $B$ is a constant. b) ${y}^{2}-y-{x}^{3}-C=0$ where $C$ is a constant. c) ${y}^{2}-y+{x}^{3}-C=0$ where $C$ is a constant. d) $y=B{e}^{{x}^{3}}+1$ where $B$ is a constant.

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
$\begin{array}{llll}\hfill \frac{dy}{dx}& =3{x}^{2}\left(y-1\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \int \frac{dy}{y-1}& =\int 3{x}^{2}\phantom{\rule{0.3em}{0ex}}dx,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill ln|y-1|& ={x}^{3}+C,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill |y-1|& =A{e}^{{x}^{3}}\phantom{\rule{1em}{0ex}}where\phantom{\rule{1em}{0ex}}A={e}^{C},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =B{e}^{{x}^{3}}+1\phantom{\rule{1em}{0ex}}where\phantom{\rule{1em}{0ex}}B\phantom{\rule{1em}{0ex}}is\phantom{\rule{1em}{0ex}}a\phantom{\rule{1em}{0ex}}constant.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

## Question 9

Using Newton’s law of cooling, it is found that the differential equation  describing the heating of a potato in an oven is
$\frac{dT}{dt}=-k\left(T-200\right)\phantom{\rule{0.3em}{0ex}}.$
What is the temperature inside the oven ?
 a) 200k${}^{o}$C b) 200${}^{o}$C c) 21.1${}^{o}$C d) The temperature cannot be determined from this equation.

Not correct. Choice (a) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

## Question 10

Use Newton’s law of cooling to determine the temperature with respect to time of a potato cooked in an oven at 200${}^{o}$C, where the temperature of the potato is 20${}^{o}$C shortly before being placed in the oven.
 a) $T=20{e}^{-kt}-200$ b) $T=180{e}^{-kt}-200$ c) $T=21-{e}^{-kt}$ d) $200-180{e}^{-kt}$

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
$\begin{array}{llll}\hfill \frac{dT}{dt}& =-k\left(T-200\right),\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \int \frac{dT}{T-200}& =\int -k\phantom{\rule{0.3em}{0ex}}dt,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill ln|T-200|& =-kt+C,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill |T-200|& =A{e}^{-kt}\phantom{\rule{1em}{0ex}}where\phantom{\rule{1em}{0ex}}A={e}^{C},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill T& =B{e}^{-kt}+200\phantom{\rule{1em}{0ex}}where\phantom{\rule{1em}{0ex}}B\phantom{\rule{1em}{0ex}}is\phantom{\rule{1em}{0ex}}a\phantom{\rule{1em}{0ex}}constant.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$ At $t=0$, $T=20\phantom{\rule{0.3em}{0ex}},$ therefore $20=B+200$ and $B=-180\phantom{\rule{0.3em}{0ex}},$ hence $T=200-180{e}^{-kt}\phantom{\rule{0.3em}{0ex}}.$