Applications of Differential Equations
We present examples where differential equations are widely applied to model natural phenomena, engineering systems and many other situations.
Application 1 : Exponential Growth - PopulationLet P(t) be a quantity that increases with time t and the rate of increase is proportional to the same quantity P as followswhere d p / d t is the first derivative of P, k > 0 and t is the time. The solution to the above first order differential equation is given by where A is a constant not equal to 0. If P = P0 at t = 0, then P0 = A e0 which gives A = P0 The final form of the solution is given by Assuming P0 is positive and since k is positive, P(t) is an increasing exponential. d P / d t = k P is also called an exponential growth model.
Application 2 : Exponential Decay - Radioactive MaterialLet M(t) be the amount of a product that decreases with time t and the rate of decrease is proportional to the amount M as followswhere d M / d t is the first derivative of M, k > 0 and t is the time. Solve the above first order differential equation to obtain where A is non zero constant. It we assume that M = M0 at t = 0, then M0 = A e0 which gives A = M0 The solution may be written as follows Assuming M0 is positive and since k is positive, M(t) is an decreasing exponential. d M / d t = - k M is also called an exponential decay model.
Application 3 : Falling ObjectAn object is dropped from a height at time t = 0. If h(t) is the height of the object at time t, a(t) the acceleration and v(t) the velocity. The relationships between a, v and h are as follows:a(t) = dv / dt , v(t) = dh / dt. For a falling object, a(t) is constant and is equal to g = -9.8 m/s. Combining the above differential equations, we can easily deduce the follwoing equation d 2h / dt 2 = g Integrate both sides of the above equation to obtain dh / dt = g t + v0 Integrate one more time to obtain h(t) = (1/2) g t2 + v0 t + h0 The above equation describes the height of a falling object, from an initial height h0 at an initial velocity v0, as a function of time.
Application 4 : Newton's Law of CoolingIt is a model that describes, mathematically, the change in temperature of an object in a given environment. The law states that the rate of change (in time) of the temperature is proportional to the difference between the temperature T of the object and the temperature Te of the environment surrounding the object.Let x = T - Te so that dx / dt = dT / dt Using the above change of variable, the above differential equation becomes The solution to the above differential equation is given by x = A e - k t substitute x by T - Te T - Te = A e - k t Assume that at t = 0 the temperature T = To To - Te = A e 0 which gives A = To - Te The final expression for T(t) i given by T(t) = Te + (To - Te)e - k t This last expression shows how the temperature T of the object changes with time.
Application 5 : RL circuit
L di/dt + Ri = E , where E is a constant voltage. Let us solve the above differential equation which may be written as follows L [ di / dt ] / [E - R i] = 1 which may be written as - (L / R) [ - R d i ] / [E - Ri] = dt Integrate both sides - (L / R) ln(E - R i) = t + c , c constant of integration. Find constant c by setting i = 0 at t = 0 (when switch is closed) which gives c = (-L / R) ln(E) Substitute c in the solution - (L / R) ln(E - R i) = t + (-L/R) ln (E) which may be written (L/R) ln (E)- (L / R) ln(E - R i) = t ln[E/(E - Ri)] = t(R/L) Change into exponential form [E/(E - Ri)] = et(R/L) Solve for i to obtain i = (E/R) (1-e-Rt/L) The starting model for the circuit is a differential equation which when solved, gives an expression of the current in the circuit as a function of time. |
More references on Differential Equations
