Category: Applications of First-Order ODE

Growth and Decay Problems

Let N(t){N}{\left({t}\right)} denote the amount of a substance (or population) that is either growing or decaying. If we assume that dNdt\frac{{{d}{N}}}{{{d}{t}}}, the time rate of change of this amount of substance, is proportional to the amount of substance present, we have that dNdt=kN\frac{{{d}{N}}}{{{d}{t}}}={k}{N}, where k{k} is the constant of proportionality.

Temperature Problems

Newton's law of cooling, which is equally applicable to heating, states that the time rate of change of the temperature of a body is proportional to the temperature difference between the body and its surrounding medium. Let T{T} denote the temperature of the body and Tm{T}_{{m}} denote the temperature of the surrounding medium. Then, the time rate of change of the temperature of the body is dTdt\frac{{{d}{T}}}{{{d}{t}}}, and Newton's law of cooling can be formulated as dTdt=k(TTm)\frac{{{d}{T}}}{{{d}{t}}}=-{k}{\left({T}-{T}_{{m}}\right)}, or as dTdt+kT=kTm\frac{{{d}{T}}}{{{d}{t}}}+{k}{T}={k}{T}_{{m}}, where k{k} is a positive constant of proportionality. Once k{k} is chosen positive, the minus sign is required in Newton's law to make dTdt\frac{{{d}{T}}}{{{d}{t}}} negative in a cooling process, when T{T} is greater than Tm{T}_{{m}}, and to make it positive in a heating process, when T{T} is smaller than Tm{T}_{{m}}.

Falling Body Problems

Consider a vertically falling body of mass m{m} that is being influenced only by gravity g{g{}} and an air resistance that is proportional to the velocity of the body. Assume that both gravity and mass remain constant and, for convenience, choose the downward direction as the positive direction.

Dilution Problems

Consider a tank which initially holds V0{V}_{{0}} gal. of brine that contains a{a} lb. of salt. Another brine solution containing b{b} lb. of salt per gallon is poured into the tank at a rate of e{e} gal./min., while, simultaneously, the well-stirred solution leaves the tank at a rate of f{f{}} gal./min. The problem is to find the amount of salt in the tank at any time t{t}. Let Q{Q} denote the amount (in pounds) of salt in the tank at any time t{t}. The time rate of change of Q{Q}, dQdt\frac{{{d}{Q}}}{{{d}{t}}}, equals the rate at which the salt enters the tank minus the rate at which the salt leaves the tank. The salt enters the tank at a rate of be{b}\cdot{e} lb./min. To determine the rate at which the salt leaves the tank, we first calculate the volume of brine in the tank at any time t{t}, which is the initial volume V0{V}_{{0}} plus the volume of brine added et{e}\cdot{t} minus the volume of brine removed ft{f{\cdot}}{t}. Thus, the volume of brine at any time is V0+etft{V}_{{0}}+{e}\cdot{t}-{f{\cdot}}{t}. The concentration of salt in the tank at any time is QV0+etft\frac{{Q}}{{{V}_{{0}}+{e}\cdot{t}-{f{\cdot}}{t}}}, from which it follows that the salt leaves the tank at a rate of f QV0+etft{f{\ }}\frac{{Q}}{{{V}_{{0}}+{e}\cdot{t}-{f{\cdot}}{t}}} gal./min.

Electrical Circuits

The basic equation governing the amount of current I{I} (in amperes) in a simple RL circuit consisting of a resistance R{R} (in ohms), an inductance L{L} (in henries), and an electromotive force (abbreviated as 'emf') E{E} (in volts) is dIdt+RLI=EL\frac{{{d}{I}}}{{{d}{t}}}+\frac{{R}}{{L}}{I}=\frac{{E}}{{L}}.For an RC circuit consisting of a resistance, a capacitance C{C} (in farads), an emf, and no inductance, the equation governing the amount of electrical charge q{q} (in coulombs) on the capacitor is dqdt+1RCq=ER\frac{{{d}{q}}}{{{d}{t}}}+\frac{{1}}{{{R}{C}}}{q}=\frac{{E}}{{R}}. The relationship between q{q} and I{I} is I=dqdt{I}=\frac{{{d}{q}}}{{{d}{t}}}.

Orthogonal Trajectories

Consider a one-parameter family of curves in the xy-plane defined by F(x,y,c)=0{F}{\left({x},{y},{c}\right)}={0}, where c{c} denotes the parameter. The problem is to find another one-parameter family of curves called the orthogonal trajectories of the family defined above and given analytically by G(x,y,k)=0{G}{\left({x},{y},{k}\right)}={0}, such that every curve in this new family intersects at right angles every curve in the original family.