# 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 - Population

Let $P(t)$ be a quantity that increases with time $t$ and the rate of increase is proportional to the same quantity $P$ as follows

$\dfrac{dP}{dt} = k P$

where $\dfrac{dP}{dt}$ 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

$P(t) = A e^{k t}$

where $A$ is a constant not equal to $0$.

If $P = P_0$ at $t = 0$, then

$P_0 = A e^0$

which gives $A = P_0$

The final form of the solution is given by

$P(t) = P_0 e^{k t}$

Assuming $P_0$ is positive and since $k$ is positive, $P(t)$ is an increasing exponential. $\dfrac{dP}{dt} = k P$ is also called an exponential growth model.

## Application 2 : Exponential Decay - Radioactive Material

Let $M(t)$ be the amount of a product that decreases with time $t$ and the rate of decrease is proportional to the amount $M$ as follows

$\dfrac{dM}{dt} = - k M$

where $\dfrac{dM}{dt}$ is the first derivative of $M$, $k > 0$ and $t$ is the time.

Solve the above first order differential equation to obtain

$M(t) = A e^{- k t}$

where $A$ is non zero constant.

It we assume that $M = M_0$ at $t = 0$, then

$M_0 = A e^0$

which gives $A = M_0$

The solution may be written as follows

$M(t) = M_0 e^{- k t}$

Assuming $M_0$ is positive and since $k$ is positive, $M(t)$ is an decreasing exponential. $\dfrac{dM}{dt} = - k M$ is also called an exponential decay model.

## Application 3 : Falling Object

An 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:

$a(t) = \dfrac{dv}{dt}$ , $v(t) = \dfrac{dh}{dt}$.

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

$\dfrac{d^2h}{dt^2}=g$

Integrate both sides of the above equation to obtain

$\dfrac{dh}{dt} = g t + v_0$

Integrate one more time to obtain

$h(t) = (1/2) g t^2 + v_0 t + h_0$

The above equation describes the height of a falling object, from an initial height $h_0$ at an initial velocity $v_0$, as a function of time.

## Application 4 : Newton's Law of Cooling

It 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.

$\dfrac{dT}{dt} = - k (T - Te)$

Let $x = T - Te$ so that $\dfrac{dx}{dt}=\dfrac{dT}{dt}$

Using the above change of variable, the above differential equation becomes

$\dfrac{dx}{dt} = - k x$

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 $T = T_0$

$T_0 - Te = A e^0$

which gives $A = T_0 - Te$

The final expression for $T(t)$ is given by

$T(t) = Te + (T_0 - Te)e^{- k t}$

This last expression shows how the temperature $T$ of the object changes with time.

## Application 5 : $RL$ circuit

Let us consider the $RL$ (resistor $R$ and inductor $L$) circuit shown above. At $t = 0$ the switch is closed and current passes through the circuit. Electricity laws state that the voltage across a resistor of resistance $R$ is equal to $R i$ and the voltage across an inductor $L$ is given by $L \dfrac{di}{dt}$($i$ is the current). Another law gives an equation relating all voltages in the above circuit as follows:

$L di/dt + R i = E$ , where $E$ is a constant voltage.

Let us solve the above differential equation which may be written as follows

$L \dfrac{di/dt}{E - R i} = 1$

which may be written as

$- \dfrac{L}{R} \dfrac{-R di}{E-R i} = dt$

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

$(L/R) \ln (E)- (L / R) \ln (E - R i) = t$

$\ln \dfrac{E}{E-R i} = t(R/L)$

Change into exponential form

$\dfrac{E}{E-R i} = e^{t(R/L)}$

Solve for $i$ to obtain

$i = \dfrac{E}{R} (1-e^{-Rt/L})$

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

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Updated: 2 April 2013

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