OBJECTIVES

When you finish this section, you should be able to:

1. Solve a first-order linear differential equation (p. 1078)
2. Solve applied problems involving first-order linear differential equations (p. 1079)
3. Find the general solution of a Bernoulli equation (p. 1082)
4. Solve applied problems (p. 1083)

NOW WORK

Problem 5.

Solving a First-Order Linear Differential Equation

Find the general solution of the first-order linear differential equation \[ \frac{dy}{dx}+\frac{x}{x^{2}+1}y=\dfrac{x^{3}}{1+x^{2}} \]

Solution Compare the differential equation to (1). Since \(P(x)= \dfrac{x}{x^{2}+1}\), the integrating factor is \[ \begin{equation*} e^{\int \!P( x)\,dx} = \exp\; \left[{\int {\frac{x}{x^{2}+1}}\,dx}\right]= \exp \left[{\frac{1}{2} \ln ( x^{2}+1) }\right] = \exp\; \Big[{\ln ( x^{2}+1) ^{1/2}}\Big]= \sqrt{x^{2}+1} \end{equation*} \]

Now we multiply the differential equation by \(\sqrt{x^{2}+1}\) and obtain \[ \begin{equation*} \sqrt{x^{2}+1}\,\frac{dy}{dx}+\dfrac{x}{\sqrt{x^{2}+1}}y=\dfrac{x^{3}}{\sqrt{ x^{2}+1}} \end{equation*} \]

The left side of the above equation is the derivative of \(\sqrt{x^{2}+1}\,y\), so we can write \[ \begin{equation*} \dfrac{d}{dx}\big( \sqrt{x^{2}+1}\,y\big) =\dfrac{x^{3}}{\sqrt{x^{2}+1}} \end{equation*} \]

Integrating both sides, we find \[ \begin{eqnarray*} \sqrt{x^{2}+1}\,y &=&\int \dfrac{x^{3}}{\sqrt{x^{2}+1}}\,dx \underset{\color{#0066A7}{\textrm{Let} \,u\,=\,x^{2}+1,\,du\,=\,2x\,dx}}{\underset{\color{#0066A7}\uparrow }{=}} \dfrac{1}{2}\int \dfrac{ \left( u-1\right) }{\sqrt{u}}\,du=\dfrac{1}{2}\int \big( {u^{1/2}}- {u^{-1/2}}\big)\,du \\ &=&\dfrac{1}{2}\left( \dfrac{u^{3/2}}{\dfrac{3}{2}}-\dfrac{u^{1/2}}{\dfrac{1 }{2}}\right) +C=\dfrac{( x^{2}+1) ^{3/2}}{3}-(x^{2}+1)^{1/2}+C \\ y &=&\dfrac{x^{2}+1}{3}-1+\dfrac{C}{\sqrt{x^{2}+1}}\qquad\color{#0066A7}{{\hbox{Solve for }y.}} \end{eqnarray*} \]

and we have the general solution to the differential equation.

NOW WORK

Problem 9.

Solving a Free Fall Problem

A parachuter and his parachute together weigh \(192 {\rm lb}\). At the instant the parachute opens, he is falling at 150 feet per second (ft/s). Suppose the air resistance exerts a force of 200 lb when the parachuter’s speed is 20 ft/s.

1. How fast is he falling \(3\) seconds after the parachute opens?
2. What is the parachuter’s limiting velocity?

Solution (a) In the U.S. customary system of units, length is measured in feet, weight in pounds, time in seconds, and \(g\approx 32\;\rm{ft}/\rm{s}^{2}\). The mass \(m\) of the parachuter and his parachute obeys \(mg=\hbox{weight}=192\;\rm{lb}\).

Since the air resistance is \(200\;\rm{lb}\) when the speed is \(20\;\rm{ft}/\rm{s}\), \[ \begin{eqnarray*} 200& =&k\cdot 20\qquad\color{#0066A7}{{F=kv}; \hbox{force is proportional to speed.}} \\ k& =&10 \end{eqnarray*} \]

From (3), the velocity of the parachuter at time \(t\) after the parachute opens is \[ \begin{eqnarray*} v(t) &=&-\frac{mg}{k}+\left( v_{0}+\frac{mg}{k}\right) e^{{{-kt/m}}} \\ v(t) &=&-\frac{192}{10}+\left( -150+\frac{192}{10}\right) e^{{{-10t/6}}}\qquad\color{#0066A7}{{v}_{0}=-150; mg=192; g=32; k=10.} \\ v(t) &=&-19.2-130.8e^{{{-5t/3}}} \end{eqnarray*} \]

After \(3\) seconds, the parachuter’s velocity is \[ \begin{equation*} v(3)=-19.2-130.8e^{-5}\approx -20.081\;\rm{ft}/\!\rm{s} (13.692 \;\rm{mi}/\!\rm{h}) \end{equation*} \]

(b) The limiting velocity is \(\lim\limits_{t\rightarrow \infty }v(t)=-\dfrac{mg}{k}=-19.2\;\rm{ft}/\rm{s} (13.091\;\rm{mi}/\rm{h} )\).

NOW WORK

Problem 47.

Solving a Flow Rate Problem

A large tank contains \(81\;\rm{gal}\) of brine in which \(20\;\rm{lb}\) of salt are dissolved. Brine containing \(3\;\rm{lb}\) of dissolved salt per gallon runs into the tank at the rate of \(5\;\rm{gal}/\rm{min}\). The mixture, kept uniform by stirring, drains from the tank at the rate of \(2 \;\rm{gal}/\rm{min}\). How much salt is in the tank at the end of \(37 \;\rm{min}\)? See Figure 6.

Figure 6

Solution Let \(y(t)\) be the amount of salt in the tank at time \(t\) . The rate of change \(\dfrac{dy}{dt}\) of salt in the tank at time \(t\) is the rate at which salt enters the tank minus the rate at which salt leaves the tank. Salt enters the tank at the rate of \(\left( 3\;\rm{lb}/\rm{gal}\right) \left(5\;\rm{gal}/\rm{min}\right) =15\;\rm{lb}/\rm{min}\).

To find the rate at which the salt exits the tank, we first need to find the concentration of salt in pounds per gallon at time \(t\). \[ \begin{equation*} \hbox{Concentration}=\dfrac{\hbox{amount of salt}}{\hbox{number of gallons}}= \frac{y(t)}{81+(5-2)t}=\dfrac{y( t) }{81+3t}\;\rm{lb}/\rm{gal} \end{equation*} \]

Then the rate at which the salt exits the tank is \[ \begin{equation*} \left( \dfrac{y( t) }{81+3t}\;\rm{lb}/\rm{gal}\right) \left( 2 \;\rm{gal}/\rm{min}\right) =\dfrac{2y( t) }{81+3t}\;\rm{lb}/\rm{min} \end{equation*} \]

The mixture problem is modeled by the differential equation \[ \begin{eqnarray*} \dfrac{dy}{dt}=\hbox{rate in }-\hbox{rate out}&=&15-\frac{2y( t) }{ 81+3t}\\ \dfrac{dy}{dt}+\frac{2}{3( 27+t) }y&=&15 \end{eqnarray*} \]

This is a first-order linear differential equation. We use the integrating factor: \[ \begin{eqnarray*} e^{\int \!P( t)\,dt}&=&\exp \!\left[ \int \!\dfrac{2}{3( 27+t) } dt\right] =\exp \!\left[ \dfrac{2}{3}\ln ( 27+t) \right]\\ &=&\exp\,[ \ln ( 27+t) ^{2/3}] =(27+t)^{2/3} \end{eqnarray*} \]

We multiply the differential equation by the integrating factor to obtain \[ \begin{eqnarray*} (27+t)^{2/3}\dfrac{dy}{dt}+\dfrac{2y}{3( 27+t) ^{1/3}} &=&15(27+t)^{2/3} \\ \frac{d}{dt}[y(t)(27+t)^{2/3}] &=&15(27+t)^{2/3} \end{eqnarray*} \]

Now we integrate both sides with respect to \(t\).

Next we use the initial condition that \(y( 0) =20\;\rm{lb}\). Then \[ \begin{eqnarray*} y( 0) &=&20=9(27)+C(27)^{-2/3}\\ -223 &=&\dfrac{C}{9} \\ C &=&-2007 \end{eqnarray*} \]

The amount of salt \(y( t) \) in the brine at time \(t\) min is \[ \begin{equation*} y(t)=9(27+t)-2007(27+t)^{-2/3} \end{equation*} \]

The amount of salt in the tank at the end of \(37\;\rm{min}\) is \[ y(37)=9(27+37)-2007(27+37)^{-2/3}\approx 451\;\rm{lb} \]

NOW WORK

Problem 49.

ORIGINS

Jakob (or James) Bernoulli (1654–1705) was a Swiss mathematician. Jakob was one of eight family members, in three generations of Bernoullis, who made major contributions to mathematics and physics.

Steps for Solving a Bernoulli Equation

\[ \dfrac{dy}{dx}+P(x)y=Q(x)y^{n}\qquad n\neq 0\quad n\neq 1 \]

Step 1 Multiply both sides of the equation by \(y^{-n}\). [This isolates \(Q( x) \) on the right.]

Step 2 Let \(v=v( x) =y^{-n+1}\). Then \(\dfrac{dv}{dx}=(1-n)y^{-n}\dfrac{dy}{dx}\).

Step 3 Substitute \(v\) and \(\dfrac{1}{1-n}\dfrac{dv}{dx}\) into \(y^{-n}\dfrac{dy}{dx}+ P(x)y^{-n+1} =Q(x)\).

Step 4 Find the general solution of the resulting first-order linear differential equation.

Step 5 Rewrite the general solution in terms of the original variables \(x\) and \(y\).

Solving a Bernoulli Equation

Find the general solution of the differential equation \[ \begin{equation*} \frac{dy}{dx}=4y+2e^{x}y^{1/2} \end{equation*} \]

Solution We begin by writing the equation in the form of (3): \[ \begin{equation*} \frac{dy}{dx}-4y=2e^{x}y^{1/2} \end{equation*} \]

This is a Bernoulli differential equation with \(P(x)=-4\), \(Q(x)=2e^{x}\), and \(n=\dfrac{1}{2}\). Then we follow the steps for solving a Bernoulli equation.

Step 1 Multiply the differential equation by \(y^{-1/2}\) to obtain \[ \begin{equation*} y^{-1/2}\frac{dy}{dx}-4y^{1/2}=2e^{x} \end{equation*} \]

Step 2 Let \(v=v( x) =y^{1/2}\). Then \(\dfrac{dv}{dx}= \dfrac{1}{2}y^{-1/2}\dfrac{dy}{dx}\).

Step 3 Substitute \(v\) and \(\dfrac{dv}{dx}\) into \(y^{-1/2}\dfrac{dy }{dx}-4y^{1/2}=2e^{x}\). \[ \begin{eqnarray*} 2\dfrac{dv}{dx}-4v& =&2e^{x} \\ \dfrac{dv}{dx}-2v& =&e^{x} \end{eqnarray*} \]

This is a first-order linear differential equation in \(x\) and \(v\).

Step 4 Multiply the differential equation by the integrating factor \(e^{\int -2\,dx}=e^{-2x}\). \[ \begin{eqnarray*} e^{-2x}\dfrac{dv}{dx}-2e^{-2x}v& =&e^{-x} \\ \dfrac{d}{dx}(e^{-2x}v)& =&e^{-x} \\ e^{-2x}v& =&\int e^{-x}dx=-e^{-x}+C\quad C > 0 \\ v& =&Ce^{2x}-e^{x} \end{eqnarray*} \]

Step 5 Using \(v=y^{1/2}\), we write the solution in terms of \(x\) and \(y\). \[ \begin{eqnarray*} y^{1/2}& =&Ce^{2x}-e^{x} \\ y& =&(Ce^{2x}-e^{x})^{2} \end{eqnarray*} \]

NOTE

Since \(e^{-2x} v + e^{-x} > 0\), then \(C > 0\).

NOW WORK

Problem 25.

ORIGINS

Daniel Bernoulli (1700–1782) was the nephew of Jakob Bernoulli and son of mathematician Daniel Bernoulli. Daniel, the son, was a medical doctor and a mathematician who did much work with Leonhard Euler (1707–1783). In 1760 Daniel Bernoulli used what is now called the logistic equation to study the spread of smallpox.

Solving a Logistic Equation

An influenza epidemic is spreading throughout a population of \(50,000\) people at a rate that is proportional to the product of the number of infected people and the number of noninfected people. Suppose that \(100\) people were infected initially and \(1000\) were infected after \(10\) days.

1. Find the number of infected people at any time \(t\). How long will it take for half the population to be infected?
2. Find the time \(t\) at which the rate of spread of infection stops increasing and begins to decrease.

Solution (a) If \(y=y( t) \) is the number of people infected at time \(t\), then \[ \begin{equation*} \dfrac{dy}{dt}=ky( 50,000-y) \tag{7} \end{equation*} \]

where \(k\) is the constant of proportionality. As given in equation (5), the solution is \[ \begin{eqnarray*} y(t)&=&\frac{100(50,000)}{100+(50,000-100) e^{-50,000kt}}\qquad \color{#0066A7}{R=100,\quad M=50,000}\\ &=&\frac{100(50,000)}{100+49,900e^{-50,000kt}} =\frac{50,000}{1+499e^{-50,000kt}} \end{eqnarray*} \]

We find \(k\) from the boundary condition \(y(10)=1000\). Then \[ \begin{eqnarray*} 1000& =&\frac{50,000}{1+499e^{-500,000k}} \\ 499e^{-500,000k}& =&49 \\ -500,000k& =&\ln \left( \frac{49}{499}\right) \\ k& \approx &0.00000464=4.64 \times 10^{-6} \end{eqnarray*} \]

So, \[ \begin{equation*} y(t)=\frac{50,000}{1+499e^{-0.232t}} \end{equation*} \]

Half the population is infected when \(y(t)=25,000\). \[ \begin{eqnarray*} 25,000& =&\dfrac{50,000}{1+499e^{-0.232t}} \\ 1+499e^{-0.232t}& =&2 \\ e^{-0.232t}& =&\frac{1}{499} \\ t& =&\frac{\ln \dfrac{1}{499}}{-0.232}\approx 27 \hbox{days} \end{eqnarray*} \]

Half the population is infected in approximately \(27\) days.

(b) The time \(t\) we seek is the inflection point of \(y=y( t) \). We need to find \(t\) so that \(y^{\prime \prime} ( t) =0\). \[ \begin{eqnarray*} y^{\prime} ( t) &=&ky( 50,000-y) =k( 50,000y-y^{2})\qquad \color{#0066A7}{{(6)}} \\ y^{\prime \prime} &=&k( 50,000y^{\prime} -2yy^{\prime} ) =0 \\ 2y &=&50,000 \\ y &=&25,000 \end{eqnarray*} \]

From (a), \(y=25,000\) when \(t=27\). That is, on Day \(27\), the rate of infection stops increasing and begins to decrease.

NOW WORK

Problem 51.