## MATH1013 Quizzes

Quiz 10: First order linear differential equations
Question 1 Questions
Which of the following first order linear differential equations are in standard form ?
 a) $x{y}^{\prime }\left(x\right)+\frac{y\left(x\right)}{x}={e}^{5x}$ b) $\frac{dy}{dx}+{x}^{2}y={x}^{3}$ c) $\frac{1}{t}\frac{dy}{dt}+y={t}^{5}$ d) ${y}^{\prime }\left(t\right)+{t}^{4}y={e}^{4t}$

There is at least one mistake.
For example, choice (a) should be False.
There is at least one mistake.
For example, choice (b) should be True.
There is at least one mistake.
For example, choice (c) should be False.
There is at least one mistake.
For example, choice (d) should be True.
Correct!
1. False
2. True
3. False
4. True
Put the first order linear differential equation
${t}^{2}\frac{dx}{dt}+\frac{x}{t}={e}^{2t}$
in standard form.
 a) $\frac{dx}{dt}+{t}^{-3}x={t}^{-2}{e}^{2t}$ b) $\frac{dx}{dt}+{t}^{-3}x={e}^{2t}$ c) ${x}^{2}\frac{dy}{dx}+\frac{y}{x}={e}^{2x}$ d) ${t}^{2}\frac{dx}{dt}+\frac{x}{t}-{e}^{2t}=0$

Choice (a) is correct!
The standard form of a differential equation, for a function $x\left(t\right)$, is ${x}^{\prime }\left(t\right)+P\left(t\right)x\left(t\right)=R\left(t\right)$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Determine which of the following could be a multiplier (integrating factor) of the first order linear differential equation
$\frac{dy}{dx}+\frac{2y}{{x}^{2}}=x.$
 a) ${e}^{-\frac{2}{3{x}^{3}}}$ b) ${e}^{-\frac{2}{x}}$ c) ${e}^{-\frac{1}{2}{x}^{2}}$ d) ${e}^{-\frac{1}{x}}$

Choice (a) is incorrect
Choice (b) is correct!
A multiplier is ${e}^{\int \frac{2}{{x}^{2}}\phantom{\rule{1em}{0ex}}dx}={e}^{-\frac{2}{x}}.$
Choice (c) is incorrect
Choice (d) is incorrect
Determine an integrating factor for the solution of the following first order linear differential equation
${t}^{3}{y}^{\prime }\left(t\right)+2y\left(t\right)={e}^{3t}.$
 a) ${e}^{2t}$ b) ${e}^{\frac{1}{3}{e}^{3t}}$ c) ${e}^{\int {t}^{-3}{e}^{3t}dt}$ d) ${e}^{-{t}^{-2}}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Putting the equation into standard form, ${y}^{\prime }\left(t\right)+2{t}^{-3}y\left(t\right)={t}^{-3}{e}^{3t}$, an integrating factor is ${e}^{\int 2{t}^{-3}dx}={e}^{-{t}^{-2}}$.
Find the general solution to the first order linear differential equation
$\frac{dy}{dx}-2y={e}^{4x}.$
 a) $y=\frac{1}{2}{e}^{2x}+C$ b) $y=\frac{1}{6}{e}^{6x}+C$ c) $y=\frac{1}{2}+C{e}^{-2x}$ d) $y=\frac{1}{2}{e}^{4x}+C{e}^{2x}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
A multiplier is ${e}^{\int -2dx}={e}^{-2x}\phantom{\rule{0.3em}{0ex}}.$ Hence our equation becomes ${e}^{-2x}\frac{dy}{dx}-2{e}^{-2x}y={e}^{2x}\phantom{\rule{0.3em}{0ex}}.$ Thus $\frac{d}{dx}\left({e}^{-2x}y\right)={e}^{2x}$ and ${e}^{-2x}y=\frac{1}{2}{e}^{2x}+C$ which gives $y=\frac{1}{2}{e}^{4x}+C{e}^{2x}$.
Find the general solution to the first order linear differential equation
$x{y}^{\prime }\left(x\right)+y\left(x\right)={x}^{3}.$
 a) $y\left(x\right)=\frac{1}{4}{x}^{3}+\frac{C}{x}$ b) $y\left(x\right)=\frac{1}{4}{x}^{4}+\frac{C}{x}$ c) $y\left(x\right)=\frac{1}{4}{x}^{3}+C$ d) $y\left(x\right){e}^{x}=\int {x}^{3}{e}^{x}dx$ and solving this is outside the scope of this course.

Choice (a) is correct!
Putting the equation in standard form we get
${y}^{\prime }\left(x\right)+\frac{y\left(x\right)}{x}={x}^{2}\phantom{\rule{0.3em}{0ex}}.$ A multiplier is thus ${e}^{\int \frac{1}{x}dx}={e}^{lnx}=x\phantom{\rule{0.3em}{0ex}}.$ Hence our equation becomes $\frac{d}{dx}\left(xy\left(x\right)\right)={x}^{3}$ which upon integrating gives $xy\left(x\right)=\frac{1}{4}{x}^{4}+C$ so
$y\left(x\right)=\frac{1}{4}{x}^{3}+\frac{C}{x}\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find the general solution to the first order linear differential equation
$\frac{dy}{dx}-2y-\frac{1}{1+{e}^{-2x}}=0\phantom{\rule{0.3em}{0ex}}.$
 a) $y=\frac{1}{2}{e}^{-2x}ln\left(1+{e}^{-2x}\right)+C{e}^{-2x}$ b) $y=-\frac{1}{2}{e}^{2x}ln\left(1+{e}^{-2x}\right)+C{e}^{2x}$ c) $y=-\frac{1}{2}{e}^{-2x}ln\left(\frac{1}{1+{e}^{-2x}}\right)+C{e}^{-2x}$ d) $y=-\frac{1}{2}{e}^{2x}ln\left(\frac{1}{1+{e}^{-2x}}\right)+C{e}^{2x}$

Choice (a) is incorrect
Choice (b) is correct!
A multiplier is ${e}^{\int -2dx}={e}^{-2x}$ hence
$y{e}^{-2x}=\int \frac{{e}^{-2x}}{1+{e}^{-2x}}\phantom{\rule{1em}{0ex}}dx\phantom{\rule{0.3em}{0ex}},$ which implies $y{e}^{-2x}=-\frac{1}{2}ln\left(1+{e}^{-2x}\right)+C$ Hence so $y=-\frac{1}{2}{e}^{2x}ln\left(1+{e}^{-2x}\right)+C{e}^{2x}\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Choice (d) is incorrect
Find the particular solution
${e}^{t}\frac{dy}{dt}+{e}^{t}ty={e}^{-{t}^{2}}\left(t+1\right).$
 a) $y={e}^{-\left({t}^{2}+t\right)}-{e}^{-\frac{1}{2}{t}^{2}}$ b) $y={e}^{-\frac{1}{2}{t}^{2}}-{e}^{-\left({t}^{2}+t\right)}$ c) $y={e}^{-\left(\frac{3}{2}{t}^{2}+t\right)}-{e}^{-\frac{1}{2}{t}^{2}}$ d) $y=1-{e}^{-\left({t}^{2}+t\right)}$

Choice (a) is incorrect
Choice (b) is correct!
Putting the equation in standard form we have $\frac{dy}{dt}+ty={e}^{-\left({t}^{2}+t\right)}\left(t+1\right)$ and a multiplier is ${e}^{\int tdt}={e}^{\frac{1}{2}{t}^{2}}\phantom{\rule{0.3em}{0ex}}.$ Hence we have $y{e}^{\frac{1}{2}{t}^{2}}=\int {e}^{-\left(\frac{1}{2}{t}^{2}+t\right)}\left(t+1\right)\phantom{\rule{1em}{0ex}}dt\phantom{\rule{0.3em}{0ex}}.$ Evaluating the integral by substitution we obtain
$y{e}^{\frac{1}{2}{t}^{2}}=-{e}^{-\left(\frac{1}{2}{t}^{2}+t\right)}+C$ so $y=-{e}^{-\left({t}^{2}+t\right)}+C{e}^{-\frac{1}{2}{t}^{2}}\phantom{\rule{0.3em}{0ex}}.$ At $t=0$ we have $y=-1+C=0⇒C=1$ and finally $y={e}^{-\frac{1}{2}{t}^{2}}-{e}^{-\left({t}^{2}+t\right)}$.
Choice (c) is incorrect
Choice (d) is incorrect
Given that $\int t{e}^{at}dt=\frac{1}{a}{e}^{at}\left(t-\frac{1}{a}\right)+C$ where $a$ is a constant, find the particular solution of
${y}^{\prime }\left(t\right)+\frac{y\left(t\right)}{t}={e}^{5t}$
if $y\left(1\right)=0\phantom{\rule{0.3em}{0ex}}.$
 a) $y\left(t\right)=\frac{{e}^{5t}}{6}-\frac{{e}^{6-t}}{6}$ b) $y\left(t\right)=\frac{{e}^{6t}}{6}\left(t-\frac{1}{6}\right)-\frac{5{e}^{6}}{36t}$ c) $y\left(t\right)=\frac{{e}^{5t}}{5t}\left(t-\frac{1}{5}\right)-\frac{4{e}^{5}}{25t}$ d) $y\left(t\right)=\frac{{e}^{5t}}{5}\left(t-\frac{1}{5}\right)+\frac{3}{50t}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
A multiplier is ${e}^{\int \frac{1}{t}dt}=t$ so $\frac{d}{dt}\left(ty\left(t\right)\right)=t{e}^{5t}$ and $ty\left(t\right)=\frac{1}{5}{e}^{5t}\left(t-\frac{1}{5}\right)+C$ hence the general solution is
$y\left(t\right)=\frac{1}{5t}{e}^{5t}\left(t-\frac{1}{5}\right)+\frac{C}{t}\phantom{\rule{0.3em}{0ex}}.$ Since $y\left(1\right)=\frac{1}{5}{e}^{5}\left(\frac{4}{5}\right)+C=0$ we have $C=-\frac{4}{25}{e}^{5}$ and then $y\left(t\right)=\frac{{e}^{5t}}{5t}\left(t-\frac{1}{5}\right)-\frac{4{e}^{5}}{25t}\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is incorrect
Find the particular solution of the first order linear differential equation
${t}^{2}\frac{dx}{dt}+tx={t}^{5}$
where $x=\frac{1}{5}$ when $t=1\phantom{\rule{0.3em}{0ex}}.$
 a) $x=\frac{{t}^{4}}{5}\phantom{\rule{0.3em}{0ex}}.$ b) $x=\frac{{t}^{5}}{4}-\frac{1}{20t}\phantom{\rule{0.3em}{0ex}}.$ c) $x=\frac{{e}^{t}}{5}\left(t-\frac{1}{5}\right)-\frac{4e}{25}\phantom{\rule{0.3em}{0ex}}.$ d) $x=\frac{{e}^{t}}{5}\left(t-\frac{1}{5}\right)-\frac{1}{t}\left(\frac{1}{5}-\frac{4e}{25}\right)\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is correct!
Putting the equation into standard form we have $\frac{dx}{dt}+\frac{1}{t}x={t}^{3}$ and a multiplier is $t$ so we have $tx=\int {t}^{4}dt=\frac{1}{5}{t}^{5}+C$ hence the general solution is
$x=\frac{{t}^{4}}{5}+\frac{C}{t}\phantom{\rule{0.3em}{0ex}}.$ Using $x=\frac{1}{5}$ when $t=1$ we have $\frac{1}{5}+\frac{C}{1}=\frac{1}{5}$ and $C=0$ so that the particular solution is $x=\frac{{t}^{4}}{5}\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect