Differential Equations/First Order Linear 2

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[edit] 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.

[edit] 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}=Ce^{kt}

Letting C=1, we get ekt


Step 2: Multiply through

ektP' + ektP = 1000ekt

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.


[edit] 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}=Ce^{kt}

Letting C=1, we get ekt


Step 2: Multiply through

ektP' + ektP = − 10tekt

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

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

Step 4: Integrate

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


[edit] 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 Ci, at a rate of Ri. We also have water leaving the tank at a concentration Co and rate Ro. We therefore have a change in concentration in the tank of

\frac{dx}{dt}=R_iC_i-R_oC_o

Thinking this through, Ri, Ci, and Ro are constants, but Co 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 V0 volume, the volume at time t is V(t) = V0 + t(ri − ro). This makes the final equation

x'=R_iC_i-\frac{R_ox}{V_0+t(R_i-R_o)}

which is an obvious linear equation. Lets solve it.

x'+\frac{R_ox}{V_0+t(R_i-R_o)}=R_iC_i

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

\int \frac{R_o}{V_0+t(R_i-R_o)}=\frac{R_oln((R_i-R_o)t+V_0)}{R_i-R_o}+C

e^{\int P(t)dt}=Ce^{\frac{R_oln((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'+((R_i-R_o)t+V_0)^{\frac{R_o}{R_i-R_o}}\frac{R_ox}{V_0+t(R_i-R_o)}=((R_i-R_o)t+V_0)^{\frac{R_o}{R_i-R_o}}R_iC_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_iC_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_iC_idt

((R_i-R_o)t+V_0)^{\frac{R_o}{R_i-R_o}}x=\frac{C_iV_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_iV_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.

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