Application of DFQ in Cooling Temperatures - Term Paper Example

The derivation of the type of equation selected provides a practical relationship in the expected areas of application. Consequently, the primary role of this project presentation is to explore the applicability of the first-order differential equation and its significant contribution to the cooling of temperatures, which is one particular area of interest as far as the subject of real-life situations, is concerned. A first-order differential equation conforms to the linearity of ordinary differential equations since the derivative part of the equation exists in the first degree (Abell & Braselton, 2004). As a result, the general representation of a first-order differential equation of linear type can be represented by the following formula.

Where dy/dx is the derivative part, P and Q are referred to as continuous functions of the variable x. in addition, X and y represent variables that are subject to manipulation. The above-mentioned formulation is the standard form of a first-order linear differential equation, thus, the derivative solutions of such equation, first take into consideration the re-writing of any equation in the standard format before working on it in terms of derivation (Abell & Braselton, 2004). Moreover, if a differential equation contains coefficients preceding the derivative part, it is recommended that the coefficients be divided throughout the equation to ensure uniformity.

U(x) dy/dx+U(x) +P(x) y= d (U(x) y/dx
U(x) y’+U(x) P(x) y=U(x) Y’+yU’(x)
P(x) =U’(x)/Ux integrates both the numerator and the denominator with respect to x.
=Natural log u(x) =integral of P(x) dx+C1
Ux=C*exp (integral Pdx) dx which is the integrating factor.
Note that we do not need the general integrating factor, therefore we assume that C =1. Through the multiplication of the original equation by the integrating factor, we obtain the following
= y’*exp (integral Pdx) dx + yP(x) exp (integral P(x) dx =Q(x)*Ux=exp {(integral(x)} dx
= d/dx[y exp integral P(x) dx =Q(x) exp {integral(x) dx thus the general derivation of this linear first order is
Y*exp {integral(x) dx =integral {exp(x) exp {integral (P(x)} dx
Beneficial features of the first order differential equations
Their derivative part only exists in the first degree. This means that dy/dx cannot exist in a form like d2y/dx2 since the latter is a derivative resulting from a second order differential equation.
When the derivative is preceded by a constant or any other variable they must be divided through the whole equation to obtain the standard form of the ordinary differential equation (Abell & Braselton, 2004).
The first-order ordinary differential equations utilize the integrating factor to obtain their solutions regardless of whether the initial value s are given or not.
The analytic solution of a first-order differential equation
The analytical solution represents the general solution of the equations and it is imperative to note that it contains arbitrary constants, which can only be calculated if there is the presence of initial value problems (Abell & Braselton, 2004).

The numerical solution of a first-order differential equation endeavors to solve a real-life problem and in this case the problem under consideration is the cooling of temperatures using the applications that are derived from these equations. Therefore, in real-life situations, Newton’s law of cooling finds applicability in the heating of substances. It thus states; the rate of change of temperature in a body is directly proportional to the difference in the temperature of the cooling body and the surrounding medium of that body (Abell & Braselton, 2004). For instance, let T represent the temperature of a cooling body
Let Tn represent the temperature of the surrounding medium then the rate of change of the temperature is represented by:
dT/dt .Thus, the association with the law is given by dT/dt=−k (T−Tn)
The negative of K is a constant of cooling, simply because cooling is a process that involves a decrease in temperature (Abell & Braselton, 2004).

Real-life situation
Consider a cooked cake that is measured at 300 degrees F. the cake is removed from an oven and then placed in a room with a temperature of 700F. If after 3 minutes the cake is found to be 2000F, determine the time it will take the cake to cool to a temperature of 1000F (Abell & Braselton, 2004).
Solution
T-Tn =Aexp-kt
A is the difference in temperature.
=200-70=230exp-kt
But time =3 min, therefore, 130=230exp-k (3) take natural logarithms on both sides
We get that k =0.1902
Substitution
100-70=230exp0.1902t
Applying natural logs again, we get the value of time t to be 10.71 minutes. Thus, comparatively the analytic solution cannot solve real-life situation problems as opposed to the latter, which can solve diverse problems as illustrated above. Read More