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3.1 Linear Models: Class Notes

3.1a

Various situations are modeled by differential equations.

• Growth and decay: ${\displaystyle \frac{dx}{dt}=kx,x(t_0)=x_0}$

If $t_0=0$, then $x_0$ is an initial population or amount.

Used for populations, radioactive decay, and drug concentrations.

${\displaystyle \frac{dx/dt}{x}=k}$: a constant relative growth/decay rate. $k$ is the relative growth/decay rate as a decimal ($k>0$ for growth, $k<0$ for decay).

• Newton’s law of heating and cooling: ${\displaystyle \frac{dT}{dt}=k(T-T_m),T(t_0)=T_0}$

$T$ is the temperature and $t$ is time. $T_m$ is the surrounding temperature (the temperature of the medium), so $T-T_m$ is the temperature difference, and $k<0$.

• Mixtures: ${\displaystyle \frac{dA}{dt}=\text{rate in}-\text{rate out},A(0)=A_0}$

• $LR$-series electrical circuit: ${\displaystyle L\frac{di}{dt}+Ri=E(t)}$

$L$: inductance (henries, h); $R$: resistance (ohms, $\Omega$); $E$: impressed voltage (volts, V); $i$: current (amperes, A).

• $RC$-series electrical circuit: ${\displaystyle R\frac{dq}{dt}+\frac{1}{C}q=E(t)}$

$C$: capacitance (farads, f); $q$: charge (coulombs, C); ${\displaystyle i=\frac{dq}{dt}}$.

Example

A population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time $t$. After 3 hours it is observed that 400 bacteria are present. After 10 hours 2000 bacteria are present. What was the initial number of bacteria?

Solution

The growth rate is proportional to the population:

$${\displaystyle \frac{dP}{dt}=kP\text{(separable, linear)}}$$

$${\displaystyle \frac{dP}{dt}-kP=0,\mu =e^{-kt}}$$

$${\displaystyle \frac{d}{dt}\left(e^{-kt}P\right)=0\Rightarrow e^{-kt}P=C\Rightarrow P=C{e}^{kt}}$$

The population at time $t=0$ is the initial population, $P_0$ (so $C=P_0$). The solution to the model is

$$P(t)=P_0{e}^{kt}.$$

Two graphs of $P(t)=P_0{e}^{kt}$, each crossing the $P$-axis at $P_0$. Left, $k>0$: the curve increases (growth). Right, $k<0$: the curve decreases toward 0 (decay).

Use the two observations, $P(3)=400$ and $P(10)=2000$:

$$2000=P_0{e}^{10k},400=P_0{e}^{3k}$$

Divide the first equation by the second ($P_0$ cancels):

$${\displaystyle 5=e^{7k}\Rightarrow k=\frac{1}{7}\ln 5\approx 0.2299}$$

The growth rate is about 23%. Then

$$400=P_0{e}^{0.2299(3)}\Rightarrow P_0\approx 200.69$$

3.1b

Example

A thermometer is taken from an inside room to the outside, where the air temperature is $5^{\circ }$ F. After 1 minute the thermometer reads ${55}^{\circ }$ F, and after 5 minutes it reads ${30}^{\circ }$ F. What is the initial temperature of the inside room?

Solution

Sketch of the temperature $T(t)$: it starts at the unknown room temperature, passes through ${55}^{\circ }$ at $t=1$ and ${30}^{\circ }$ at $t=5$, and decreases toward the horizontal asymptote $T_m=5$. The initial gap between $T$ and $T_m$ is $D_0$.

$${\displaystyle \frac{dT}{dt}=k(T-T_m)\text{(separable, linear)}}$$

$${\displaystyle \frac{dT}{T-T_m}=kdt\Rightarrow \ln |T-T_m|=kt+C\Rightarrow T-T_m=D_0{e}^{kt}}$$

$$T(t)=T_m+D_0{e}^{kt}$$

$D_0$ is the initial temperature difference: $D_0=T_0-T_m$.

Left, cooling: $T$ starts $D_0$ degrees above $T_m$ and decreases toward $T_m$ (here $D_0>0$). Right, heating: $T$ starts below $T_m$ and increases toward it (here $D_0<0$). In both, $k<0$.

Here $T_m=5$, so

$$T(t)=5+D_0{e}^{kt}.$$

Use $T(1)=55$ and $T(5)=30$:

$$55=5+D_0{e}^k\Rightarrow 50=D_0{e}^k$$

$$30=5+D_0{e}^{5k}\Rightarrow 25=D_0{e}^{5k}$$

Dividing gives

$${\displaystyle 2=e^{-4k}\Rightarrow k=\frac{1}{4}\ln \frac{1}{2}=-\frac{1}{4}\ln 2\approx -0.1733}$$

Then, from $50=D_0{e}^k$:

$$D_0=50e^{0.1733}\approx 59.5$$

Since $D_0=T_0-T_m$ and $T_m=5$, the initial thermometer temperature is ${59.5}^{\circ }+5^{\circ }$.

3.1c

Example

A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into the tank at a rate of 5 gal/min. The well-mixed solution is pumped out at the same rate. Find the number $A(t)$ of pounds of salt in the tank at time $t$.

Solution

The tank starts with pure water, so $A(0)=0$.

A tank initially holding 500 gallons of pure water. Brine with 2 lb of salt per gallon flows in at 5 gal/min; the well-mixed solution flows out at 5 gal/min, so the volume stays 500 gallons.

Rate salt in: ${\displaystyle 2\frac{\text{lb}}{\text{gal}}·5\frac{\text{gal}}{\text{min}}=10\frac{\text{lb}}{\text{min}}}$

Rate salt out: ${\displaystyle \frac{A\text{ lb}}{500\text{ gal}}·5\frac{\text{gal}}{\text{min}}=\frac{A}{100}\frac{\text{lb}}{\text{min}}}$

$${\displaystyle \frac{dA}{dt}=\text{rate in}-\text{rate out}=10-\frac{1}{100}A,A(0)=0}$$

This is linear:

$${\displaystyle \frac{dA}{dt}+\frac{1}{100}A=10,\mu =e^{t/100}}$$

$${\displaystyle \frac{d}{dt}\left[e^{t/100}A\right]=10e^{t/100}}$$

$$e^{t/100}A=1000e^{t/100}+C$$

$$A(t)=1000+C{e}^{-t/100}$$

$A(0)=0$ gives $C=-1000$.

In the long run, ${\displaystyle \underset{t\to \infty }{lim}A(t)=1000}$ lb.

Example

Solve the previous problem under the assumption that the solution is pumped out at a faster rate of 10 gal/min. When is the tank empty?

Solution

Still $A(0)=0$ with a 500-gallon tank and salt coming in at 10 lb/min. The tank gains 5 gal/min and loses 10 gal/min, so it is losing 5 gal/min: the volume at time $t$ is $500-5t$ gallons, and the tank is empty in 100 min.

A tank initially holding 500 gallons of pure water. Brine with 2 lb of salt per gallon flows in at 5 gal/min; the well-mixed solution flows out at 10 gal/min, so the volume at time $t$ is $500-5t$ gallons.

Rate salt out: ${\displaystyle \frac{A\text{ lb}}{(500-5t)\text{ gal}}·10\frac{\text{gal}}{\text{min}}=\frac{2A}{100-t}\frac{\text{lb}}{\text{min}}}$

$${\displaystyle \frac{dA}{dt}=10-\frac{2A}{100-t},A(0)=0}$$

$${\displaystyle \frac{dA}{dt}+\frac{2}{100-t}A=10}$$

$${\displaystyle \mu =e^{\int \frac{2}{100-t}dt}=e^{-2\ln |100-t|}=\frac{1}{(100-t{)}^2}}$$

$${\displaystyle \frac{d}{dt}\left[\frac{1}{(100-t{)}^2}A\right]=\frac{10}{(100-t{)}^2}}$$

$${\displaystyle \frac{1}{(100-t{)}^2}A=\frac{10}{100-t}+C}$$

$$A=10(100-t)+C(100-t{)}^2$$

$A(0)=0$: $0=1000+10000C$, so ${\displaystyle C=-\frac{1}{10}}$.

The tank is empty at $t=100$ min.

Example

A 200-volt electromotive force is applied to an $RC$-series circuit in which the resistance is 1000 ohms and the capacitance is $5\times {10}^{-6}$ farad. Find the charge $q(t)$ on the capacitor if $i(0)=0.4$. Determine the charge and current at $t=0.005$ s. Determine the charge as $t\to \infty$.

Solution

$${\displaystyle R\frac{dq}{dt}+\frac{1}{C}q=E(t),i=\frac{dq}{dt}}$$

Here $R=1000$, $E(t)=200$, and ${\displaystyle C=5\times {10}^{-6}=\frac{5}{{10}^6}}$, so ${\displaystyle \frac{1}{C}=\frac{{10}^6}{5}}$.

$${\displaystyle 1000\frac{dq}{dt}+\frac{{10}^6}{5}q=200}$$

Divide by 1000:

$${\displaystyle \frac{dq}{dt}+200q=\frac{1}{5},\mu =e^{200t}}$$

$${\displaystyle \frac{d}{dt}\left(e^{200t}q\right)=\frac{1}{5}{e}^{200t}}$$

$${\displaystyle e^{200t}q=\frac{1}{1000}{e}^{200t}+c}$$

$${\displaystyle q(t)=\frac{1}{1000}+c{e}^{-200t}}$$

The initial condition is on the current, ${\displaystyle i=\frac{dq}{dt}}$:

$$i(t)=-200c{e}^{-200t}$$

$${\displaystyle i(0)=0.4\Rightarrow 0.4=-200c\Rightarrow c=-\frac{0.4}{200}=-\frac{1}{500}}$$

At $t=0.005$ s (so $200t=1$):

$${\displaystyle q(0.005)=\frac{1}{1000}-\frac{1}{500}{e}^{-1}\approx 0.000264\text{ coulombs}}$$

$$i(0.005)=0.4e^{-1}\approx 0.1472\text{ amps}$$

As $t\to \infty$, ${\displaystyle q\to \frac{1}{1000}}$ coulomb.