# List of Notes - Category: Applications of First-Order ODE

## Growth and Decay Problems

Let N(t) denote amount of substance (or population) that is either growing or decaying. If we assume that `(dN)/(dt)` , the time rate of change of this amount of substance, is proportional to the amount of substance present, then `(dN)/(dt)=kN` where 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 denote the temperature of the body and let `T_m` denote the temperature of the surrounding medium. Then the time rate of change of the temperature of the body is `(dT)/(dt)` , and Newton's law of cooling can be formulated as `(dT)/(dt)=-k(T-T_m)` or as `(dT)/(dt)+kT=kT_m` where k is a positive constant of proportionality. Once k is chosen positive, the minus sign is required in Newton's law to make `(dT)/(dt)` negative in a cooling process, when T is greater than `T_m`, and positive in a heating process, when T is less than `T_m`.

## Falling Body Problems

Consider a vertically falling body of mass m that is being influenced only by gravity 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.Newton's second law of motion: The net force acting on a body is equal to the time rate of change of the momentum of the body; or, for constant mass, `F=m(dv)/(dt)` . For the problem at hand, there are two forces acting on the body: the force due to gravity given by the weight w of the body, which equals mg, and the force due to air resistance given by —kv, where k > 0 is a constant of proportionality. The minus sign is required because this force opposes the velocity; that is, it acts in the upward, or negative, direction. The net force F on the body is, therefore, F = mg-kv. Substituting this result we obtain `mg-kv=m(dv)/(dt)` or `(dv)/(dt)+k/m v=g` as the equation of motion for the body.If air resistance is negligible or nonexistent, then k = 0 and equation simplifies to `(dv)/(dt)=g` .When k > 0, the limiting velocity `v_l` is defined by `v_l=(mg)/k` .

## Dilution Problems

Consider a tank which initially holds `V_0` gal of brine that contains a lb of salt. Another brine solution, containing b lb of salt per gallon, is poured into the tank at the rate of e gal/min while, simultaneously, the well-stirred solution leaves the tank at the rate of f gal/min. The problem is to find the amount of salt in the tank at any time t. Let Q denote the amount (in pounds) of salt in the tank at any time t. The time rate of change of Q, `(dQ)/(dt)` , equals the rate at which salt enters the tank minus the rate at which salt leaves the tank. Salt enters the tank at the rate of be lb/min. To determine the rate at which salt leaves the tank, we first calculate the volume of brine in the tank at any time t, which is the initial volume `V_0` plus the volume of brine added et minus the volume of brine removed ft . Thus, the volume of brine at any time is `V_0+et-ft` . The concentration of salt in the tank at any time is `Q/(V_0+et-ft)` , from which it follows that salt leaves the tank at the rate of `f\ Q/(V_0+et-ft)` gal/min.

## Electrical Circuits

The basic equation governing the amount of current I (in amperes) in a simple RL circuit consisting of a resistance R (in ohms), an inductor L (in henries), and an electromotive force (abbreviated emf) E (in volts) is `(dI)/(dt)+R/L I=E/L` .For an RC circuit consisting of a resistance, a capacitance C (in farads), an emf, and no inductance, the equation governing the amount of electrical charge q (in coulombs) on the capacitor is `(dq)/(dt)+1/(RC) q=E/R` .The relationship between q and I is `I=(dq)/(dt)` .

## Orthogonal Trajectories

Consider a one-parameter family of curves in the xy-plane defined by F(x,y,c)=0 where 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 such that every curve in this new family intersects at right angles every curve in the original family.