DiffEq Home

Modeling with First Order ODEs

Radioactive Decay: Let $$N(t)$$ be the mass of a radioactive element at time $$t$$. The rate of change (decay) of $$N$$ is proportional to its current value. $\frac{dN}{dt}=-kN,$ where $$k > 0$$ depends on the element. Solving this ODE we get $$N=ce^{-kt}$$. Suppose the initial amount is $$N_0=N(0)$$. Then we get $$c=N_0$$. Thus the solution is $N(t)=N_0e^{-kt}.$ Carbon Dating: There are 2 types of carbon atoms: $$^{12}C$$ (the stable nuclide) and radioactive $$^{14}C$$ with a halflife of about 5,730 years. The ratio $$^{14}C: ^{12}C$$ is approximately constant in nature. A living creature taking carbon from nature is made up with this ratio. After its death $$^{14}C$$ starts to decay.

Example. Suppose $$25\%$$ of the original amount of $$^{14}C$$ remained in a fossil. Find the age of the fossil.

Solution. First we find $$k$$ for $$^{14}C$$ in $$N(t)=N_0e^{-kt}$$. We know that $$\frac{N_0}{2}=N_0e^{-5730k}$$. Solving we get $$k=\ln 2/5730$$ and hence $N(t)=N_0e^{-t\ln 2/5730}.$ For the fossil we want to find $$t$$ for which $$N(t)=25N_0/100$$. Plugging this into the preceding equation we get $25N_0/100=N_0e^{-t\ln 2/5730} \implies t=11,460 \text{ years}$ The answer is not surprising because of the halflife of $$^{14}C$$ (about 5,730 years).

Newton's Law of Cooling: Let $$T(t)$$ be the temperature of an object at time $$t$$. Then $\frac{dT}{dt}=-k(T-T_a),$ where $$k > 0$$ is a constant and $$T_a$$ is the constant ambient temperature. Solving this ODE we get $$T=T_a+ce^{-kt}$$. Suppose the initial temperature is $$T_0=T(0)$$. Then we get $$c=T_0-T_a$$. Thus the solution is $T(t)=T_a+(T_0-T_a)e^{-kt}.$ Note that $$T(t)=T_a+(T_0-T_a)e^{-kt}\to T_a$$ as $$t\to \infty$$.

Kirchhoff's Circuit Law: Consider an electric circuit with a capacitor, resistor, and battery. Let $$Q(t)$$ be the charge of the capacitor at time $$t$$. Then $R\frac{dQ}{dt}+\frac{Q}{C}=V,$ where $$R$$ is the constant resistant, $$C$$ is the constant capacitance, and $$V$$ is the constant voltage supplied. Solving this ODE we get $$Q=CV+ke^{-t/RC}$$. Suppose the initial charge is $$Q(0)=0$$. Then we get $$k=-CV$$. Thus the solution is $Q(t)=CV(1-e^{-t/RC}).$ Note that $$Q(t)=CV(1-e^{-t/RC})\to CV$$ as $$t\to \infty$$.

Mixing Rate Problems: The key idea for mixing rate problems is that the effective rate is the rate in minus the rate out.

Example. Suppose a tank contains 200 gal of salt water containing 100 lb salt. Then salt water with 3 lb salt/gal is pumped into the tank at a rate $$2$$ gal/min and the "well-stirred" salt water is pumped out of the tank at a rate $$2$$ gal/min. Suppose $$Q(t)$$ denotes the amount of salt in lb at time $$t$$ min. Then $\frac{dQ}{dt}=2\cdot 3-2\cdot \frac{Q}{200},\, Q(0)=100$ Solving this ODE we get $$Q=600+ce^{-t/100}$$. Using the initial condition $$Q(0)=100$$ we get $$c=-500$$. Thus the solution is $Q(t)=600-500e^{-t/100}.$ Note that $$Q(t)=600-500e^{-t/100}\to 600$$ as $$t\to \infty$$.

Last edited