Ordinary Differential Equations/First Order Linear 2

Return to Exponential Growth

Remember the population growth problem, where $\frac{dP}{dt}=(B-D)P$? Now that we can solve linear equations, we can also solve variations where a factor $f(t)$ is added in. The new equation is $\frac{dP}{dt}=(B-D)P+f(t)$, and can be solved by the linear methods taught in the last section.

Immigration

Lets say that 1000 people move into a city, in addition to the normal population growth. This can be interpreted by making $f(x)=1000$. This gives us a linear differential equation to solve

$\frac{dP}{dt}=kP+1000$

$\frac{dP}{dt}-kP=1000$

Step 1: Find $e^{\int P(t)dt}$

$\int kdt=kt+C$

$e^{\int P(t)dt}=C{e}^{kt}$

Letting C=1, we get $e^{kt}$

Step 2: Multiply through

$e^{kt}{P}^{\prime }+e^{kt}P=1000e^{kt}$

Step 3: Recognize that the left hand is $\frac{d}{dt}{e}^{\int P(t)dt}y$

$\frac{d}{dt}{e}^{kt}P=1000e^{kt}$

Step 4: Integrate

$\int (\frac{d}{dt}{e}^{kt}P)dt=\int 1000e^{kt}dt$

$e^{kt}P=\frac{1000}{k}{e}^{mt}+C$

Step 5: Solve for y

$P=\frac{1000}{k}+\frac{C}{e^{kt}}$

See how the answer is a constant addition to the normal solution, as expected.

Hunting

Lets say the government allows 10 animals to be killed a year. This makes $f(t)=-10t$. How does this effect the solution?

$\frac{dP}{dt}=kP-10t$

$\frac{dP}{dt}-kP=-10t$

Step 1: Find $e^{\int P(t)dt}$

$\int kdt=kt+C$

$e^{\int P(t)dt}=C{e}^{kt}$

Letting C=1, we get $e^{kt}$

Step 2: Multiply through

$e^{kt}{P}^{\prime }+e^{kt}P=-10t{e}^{kt}$

Step 3: Recognize that the left hand is $\frac{d}{dt}{e}^{\int P(t)dt}y$

$\frac{d}{dt}{e}^{kt}P=-10t{e}^{kt}$

Step 4: Integrate

$\int (\frac{d}{dt}{e}^{kt}P)dt=\int -10t{e}^{kt}dt$

$e^{kt}P=\frac{-10(kx-1)e^{kx}}{k^2}+C$

Step 5: Solve for y

$P=\frac{-10(kx-1)}{k^2}+\frac{C}{e^{kx}}$

Mixture problems

Imagine we have a tank containing a solution of water and some other substance (say salt). We have water coming into the tank with a concentration $C_i$, at a rate of $R_i$. We also have water leaving the tank at a concentration $C_o$ and rate $R_o$. We therefore have a change in concentration in the tank of

$\frac{dx}{dt}=R_i{C}_i-R_o{C}_o$

Thinking this through, $R_i$, $C_i$, and $R_o$ are constants, but $C_o$ depends on the current concentration of the tank, which is not constant. The current concentration is $\frac{x}{V}$ where V is the volume of water in the tank. Unfortunately, the volume is changing based on how much water is in the tank. If the tank initially has $V_0$ volume, the volume at time t is $V(t)=V_0+t(r_i-r_o)$. This makes the final equation

$x^{\prime }=R_i{C}_i-\frac{R_{o}x}{V_0+t(R_i-R_o)}$

which is an obvious linear equation. Lets solve it.

$x^{\prime }+\frac{R_{o}x}{V_0+t(R_i-R_o)}=R_i{C}_i$

Step 1: Find $e^{\int P(t)dt}$

$\int \frac{R_o}{V_0+t(R_i-R_o)}=\frac{R_{o}ln((R_i-R_o)t+V_0)}{R_i-R_o}+C$

$e^{\int P(t)dt}=C{e}^{\frac{R_{o}ln((R_i-R_o)t+V_0)}{R_i-R_o}}=C((R_i-R_o)t+V_0{)}^{\frac{R_o}{R_i-R_o}}$

Letting C=1, we get $((R_i-R_o)t+V_0{)}^{\frac{R_o}{R_i-R_o}}$

Step 2: Multiply through

$((R_i-R_o)t+V_0{)}^{\frac{R_o}{R_i-R_o}}{x}^{\prime }+((R_i-R_o)t+V_0{)}^{\frac{R_o}{R_i-R_o}}\frac{R_{o}x}{V_0+t(R_i-R_o)}=((R_i-R_o)t+V_0{)}^{\frac{R_o}{R_i-R_o}}{R}_i{C}_i$

Step 3: Recognize that the left hand is $\frac{d}{dt}{e}^{\int P(t)dt}x$

$\frac{d}{dt}((R_i-R_o)t+V_0{)}^{\frac{R_o}{R_i-R_o}}x=((R_i-R_o)t+V_0{)}^{\frac{R_o}{R_i-R_o}}{R}_i{C}_i$

Step 4: Integrate

$\int (((R_i-R_o)t+V_0{)}^{\frac{R_o}{R_i-R_o}}x)dt=\int (R_i-R_o)t+V_0{)}^{\frac{R_o}{R_i-R_o}}{R}_i{C}_{i}dt$

$((R_i-R_o)t+V_0{)}^{\frac{R_o}{R_i-R_o}}x=\frac{C_i{V}_0((R_i-R_o)t+V_0{)}^{\frac{R_i}{V_0}}}{(R_i-R_o)}$

Step 5: Solve for y

$x=\frac{C_i{V}_0((R_i-R_o)t+V_0{)}^{\frac{R_i}{V_0}}}{(R_i-R_o)((R_i-R_o)t+V_0{)}^{\frac{R_o}{R_i-R_o}}}$

Ugly, isn't it. Most of the time when dealing with real world mixture problems, you'll plug in much earlier and use numbers, which makes it easier to deal with.