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

$$\int kdt=kt+C$$

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

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

Step 2: Multiply through

$$e^{kt}P'+e^{kt}P=1000e^{kt}$$

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

$$\int kdt=kt+C$$

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

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

Step 2: Multiply through

$$e^{kt}P'+e^{kt}P=-10te^{kt}$$

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

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

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

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

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

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

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.