# 4.5 First-order linear equations  (Page 6/10)

 Page 6 / 10

A circuit has in series an electromotive force given by $E=20\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}5t$ V, a capacitor with capacitance $0.02\phantom{\rule{0.2em}{0ex}}\text{F},$ and a resistor of $8\phantom{\rule{0.2em}{0ex}}\text{Ω}.$ If the initial charge is $4\phantom{\rule{0.2em}{0ex}}\text{C},$ find the charge at time $t>0.$

Initial-value problem:

$8{q}^{\prime }+\frac{1}{0.02}q=20\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}5t,\phantom{\rule{1em}{0ex}}q\left(0\right)=4$

$q\left(t\right)=\frac{10\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}5t-8\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}5t+172{e}^{-6.25t}}{41}$

## Key concepts

• Any first-order linear differential equation can be written in the form $y\prime +p\left(x\right)y=q\left(x\right).$
• We can use a five-step problem-solving strategy for solving a first-order linear differential equation that may or may not include an initial value.
• Applications of first-order linear differential equations include determining motion of a rising or falling object with air resistance and finding current in an electrical circuit.

## Key equations

• standard form
$y\prime +p\left(x\right)y=q\left(x\right)$
• integrating factor
$\mu \left(x\right)={e}^{\int p\left(x\right)\phantom{\rule{0.1em}{0ex}}dx}$

Are the following differential equations linear? Explain your reasoning.

$\frac{dy}{dx}={x}^{2}y+\text{sin}\phantom{\rule{0.1em}{0ex}}x$

$\frac{dy}{dt}=ty$

Yes

$\frac{dy}{dt}+{y}^{2}=x$

$y\prime ={x}^{3}+{e}^{x}$

Yes

$y\prime =y+{e}^{y}$

Write the following first-order differential equations in standard form.

$y\prime ={x}^{3}y+\text{sin}\phantom{\rule{0.1em}{0ex}}x$

$y\prime -{x}^{3}y=\text{sin}\phantom{\rule{0.1em}{0ex}}x$

$y\prime +3y-\text{ln}\phantom{\rule{0.1em}{0ex}}x=0$

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

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

$\frac{dy}{dt}=4y+ty+\text{tan}\phantom{\rule{0.1em}{0ex}}t$

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

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

What are the integrating factors for the following differential equations?

$y\prime =xy+3$

$y\prime +{e}^{x}y=\text{sin}\phantom{\rule{0.1em}{0ex}}x$

${e}^{x}$

$y\prime =x\phantom{\rule{0.1em}{0ex}}\text{ln}\left(x\right)y+3x$

$\frac{dy}{dx}=\text{tanh}\left(x\right)y+1$

$\text{−}\text{ln}\left(\text{cosh}\phantom{\rule{0.1em}{0ex}}x\right)$

$\frac{dy}{dt}+3ty={e}^{t}y$

Solve the following differential equations by using integrating factors.

$y\prime =3y+2$

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

$y\prime =2y-{x}^{2}$

$xy\prime =3y-6{x}^{2}$

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

$\left(x+2\right)y\prime =3x+y$

$y\prime =3x+xy$

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

$xy\prime =x+y$

$\text{sin}\left(x\right)y\prime =y+2x$

$y=C\phantom{\rule{0.1em}{0ex}}\text{tan}\left(\frac{x}{2}\right)-2x+4\phantom{\rule{0.1em}{0ex}}\text{tan}\left(\frac{x}{2}\right)\text{ln}\left(\text{sin}\left(\frac{x}{2}\right)\right)$

$y\prime =y+{e}^{x}$

$xy\prime =3y+{x}^{2}$

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

$y\prime +\text{ln}\phantom{\rule{0.1em}{0ex}}x=\frac{y}{x}$

Solve the following differential equations. Use your calculator to draw a family of solutions. Are there certain initial conditions that change the behavior of the solution?

[T] $\left(x+2\right)y\prime =2y-1$

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

[T] $y\prime =3{e}^{t\text{/}3}-2y$

[T] $xy\prime +\frac{y}{2}=\text{sin}\left(3t\right)$

$y=\frac{C}{\sqrt{x}}+2\phantom{\rule{0.1em}{0ex}}\text{sin}\left(3t\right)$

[T] $xy\prime =2\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x}{x}-3y$

[T] $\left(x+1\right)y\prime =3y+{x}^{2}+2x+1$

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

[T] $\text{sin}\left(x\right)y\prime +\text{cos}\left(x\right)y=2x$

[T] $\sqrt{{x}^{2}+1}y\prime =y+2$

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

[T] ${x}^{3}y\prime +2{x}^{2}y=x+1$

Solve the following initial-value problems by using integrating factors.

$y\prime +y=x,y\left(0\right)=3$

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

$y\prime =y+2{x}^{2},y\left(0\right)=0$

$xy\prime =y-3{x}^{3},y\left(1\right)=0$

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

${x}^{2}y\prime =xy-\text{ln}\phantom{\rule{0.1em}{0ex}}x,y\left(1\right)=1$

$\left(1+{x}^{2}\right)y\prime =y-1,y\left(0\right)=0$

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

$xy\prime =y+2x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x,y\left(1\right)=5$

$\left(2+x\right)y\prime =y+2+x,y\left(0\right)=0$

$y=\left(x+2\right)\text{ln}\phantom{\rule{0.1em}{0ex}}\left(\frac{x+2}{2}\right)$

$y\prime =xy+2x{e}^{x},y\left(0\right)=2$

$\sqrt{x}y\prime =y+2x,y\left(0\right)=1$

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

$y\prime =2y+x{e}^{x},y\left(0\right)=-1$

A falling object of mass $m$ can reach terminal velocity when the drag force is proportional to its velocity, with proportionality constant $k.$ Set up the differential equation and solve for the velocity given an initial velocity of $0.$

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

Using your expression from the preceding problem, what is the terminal velocity? ( Hint: Examine the limiting behavior; does the velocity approach a value?)

[T] Using your equation for terminal velocity, solve for the distance fallen. How long does it take to fall $5000$ meters if the mass is $100$ kilograms, the acceleration due to gravity is $9.8$ m/s 2 and the proportionality constant is $4?$

$40.451$ seconds

A more accurate way to describe terminal velocity is that the drag force is proportional to the square of velocity, with a proportionality constant $k.$ Set up the differential equation and solve for the velocity.

Using your expression from the preceding problem, what is the terminal velocity? ( Hint: Examine the limiting behavior: Does the velocity approach a value?)

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

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