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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 $8\phantom{\rule{0.2em}{0ex}}\text{\Omega}.$ If the initial charge is $4\phantom{\rule{0.2em}{0ex}}\text{C},$ find the charge at time $t>0.$
Initial-value problem:
$8{q}^{\prime}+\frac{1}{0.02}q=20\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}5t,\phantom{\rule{1em}{0ex}}q\left(0\right)=4$
$q\left(t\right)=\frac{10\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}5t-8\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}5t+172{e}^{\mathrm{-6.25}t}}{41}$
Are the following differential equations linear? Explain your reasoning.
$\frac{dy}{dx}={x}^{2}y+\text{sin}\phantom{\rule{0.1em}{0ex}}x$
$\frac{dy}{dt}+{y}^{2}=x$
$y\prime =y+{e}^{y}$
Write the following first-order differential equations in standard form.
$y\prime ={x}^{3}y+\text{sin}\phantom{\rule{0.1em}{0ex}}x$
$y\prime -{x}^{3}y=\text{sin}\phantom{\rule{0.1em}{0ex}}x$
$y\prime +3y-\text{ln}\phantom{\rule{0.1em}{0ex}}x=0$
$\text{\u2212}xy\prime =\left(3x+2\right)y+x{e}^{x}$
$y\prime +\frac{\left(3x+2\right)}{x}y=\text{\u2212}{e}^{x}$
$\frac{dy}{dt}=4y+ty+\text{tan}\phantom{\rule{0.1em}{0ex}}t$
$\frac{dy}{dt}=yx\left(x+1\right)$
$\frac{dy}{dt}-yx\left(x+1\right)=0$
What are the integrating factors for the following differential equations?
$y\prime =xy+3$
$y\prime +{e}^{x}y=\text{sin}\phantom{\rule{0.1em}{0ex}}x$
${e}^{x}$
$y\prime =x\phantom{\rule{0.1em}{0ex}}\text{ln}\left(x\right)y+3x$
$\frac{dy}{dx}=\text{tanh}\left(x\right)y+1$
$\text{\u2212}\text{ln}\left(\text{cosh}\phantom{\rule{0.1em}{0ex}}x\right)$
$\frac{dy}{dt}+3ty={e}^{t}y$
Solve the following differential equations by using integrating factors.
$y\prime =2y-{x}^{2}$
$\left(x+2\right)y\prime =3x+y$
$xy\prime =x+y$
$\text{sin}(x)y\prime =y+2x$
$y=C\phantom{\rule{0.1em}{0ex}}\text{tan}\left(\frac{x}{2}\right)-2x+4\phantom{\rule{0.1em}{0ex}}\text{tan}\left(\frac{x}{2}\right)\text{ln}\left(\text{sin}\left(\frac{x}{2}\right)\right)$
$y\prime =y+{e}^{x}$
$y\prime +\text{ln}\phantom{\rule{0.1em}{0ex}}x=\frac{y}{x}$
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] $y\prime =3{e}^{t\text{/}3}-2y$
[T] $xy\prime +\frac{y}{2}=\text{sin}(3t)$
$y=\frac{C}{\sqrt{x}}+2\phantom{\rule{0.1em}{0ex}}\text{sin}(3t)$
[T] $xy\prime =2\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x}{x}-3y$
[T] $(x+1)y\prime =3y+{x}^{2}+2x+1$
$y=C{(x+1)}^{3}-{x}^{2}-2x-1$
[T] $\text{sin}(x)y\prime +\text{cos}(x)y=2x$
[T] $\sqrt{{x}^{2}+1}y\prime =y+2$
$y=C{e}^{{\text{sinh}}^{\mathrm{-1}}x}-2$
[T] ${x}^{3}y\prime +2{x}^{2}y=x+1$
Solve the following initial-value problems by using integrating factors.
$y\prime =y+2{x}^{2},y(0)=0$
$xy\prime =y-3{x}^{3},y(1)=0$
$y=-\frac{3x}{2}\left({x}^{2}-1\right)$
${x}^{2}y\prime =xy-\text{ln}\phantom{\rule{0.1em}{0ex}}x,y(1)=1$
$\left(1+{x}^{2}\right)y\prime =y-1,y(0)=0$
$y=1-{e}^{{\text{tan}}^{\mathrm{-1}}x}$
$xy\prime =y+2x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x,y(1)=5$
$(2+x)y\prime =y+2+x,y(0)=0$
$y=(x+2)\text{ln}\phantom{\rule{0.1em}{0ex}}\left(\frac{x+2}{2}\right)$
$y\prime =xy+2x{e}^{x},y(0)=2$
$\sqrt{x}y\prime =y+2x,y(0)=1$
$y=2{e}^{2\sqrt{x}}-2x-2\sqrt{x}-1$
$y\prime =2y+x{e}^{x},y(0)=\mathrm{-1}$
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 $0.$
$v(t)=\frac{gm}{k}\left(1-{e}^{\text{\u2212}kt\text{/}m}\right)$
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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