Differential Equations and Modelling

A differential equation is an equation where a derivative is present. For example, $\frac{dy}{dx}=3x^2$ is a differential equation, as is $2x+y^2\frac{dy}{dx}=5$ . An exam question may ask you to solve a differential equation (see below) or it may ask you to construct one from some given information or both. To construct a differential equation we first recall the rules for forming equations using proportionality:

if $y$ is directly proportional to $x$ , we write $y=kx$ , where $k$ is the constant of proportionality that can be found given more information
or if $y$ is inversely proportional to $x$ , then we would write $y=\frac{k}{x}$

More often than not, the information will include more complicated functions of other variables. For example, if $t$ is directly proportional to the square root of $s$ , you should write the equation $t=k\sqrt{s}$ . Or if $p$ is inversely proportional to the cube of $q$ then write $p=\frac{k}{q^3}$ . Consider the differential equation that describes radioactive decay. The rate at which radioactive particles decay is proportional to the number of particles that remain. Let $N(t)$ be the number of particles at any time $t$ . The rate at which it changes with respect to time is written $\frac{dN}{dt}$ . This rate is proportional to $N$ and so we can write $\frac{dN}{dt}=-kN$ where $k$ is the constant of proportionality. Note that $N$ is going down when particles decay and so the gradient/rate must be negative, hence inclusion of the minus sign. See Example 1.

Related Rates

As in the radioactive decay equation above, rates are obtained when variables are differentiated with respect to $t$ . Related rates are those that we can obtain by introducing another variable. For example, consider the area $A$ of a circle found from its radius $r$ : $A=\pi r^2$ . If the radius changes with respect to time then the derivative $\frac{dr}{dt}$ describes the rate at which this happens. If the radius changes with respect to time so does the area and the derivative $\frac{dA}{dt}$ describes the rate at which this happens. By the chain rule, these rates are related via the equation $\frac{dA}{dt}=\frac{dA}{dr}\frac{dr}{dt}$ . Since we know that $A=\pi r^2$ , we have $\frac{dA}{dr}=2\pi r$ . It follows that $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$ . See Example 2.

Solving Differential Equations

At this level, there are two ways to solve a differential equation. Consider the differential equation $\frac{dy}{dx}=3x^2$ . We can solve this by simply integrating both sides with respect to $x$ to obtain $y=x^3+c$ . Given a little more information, it may be possible to find the constant of integration – see Example 1. However, many differential equations are not stated so simply. The solution of many differential equations will require a process called separation of variables. The idea is to transform the original differential equation of the form $\frac{dy}{dx}=f(x,y)$ into the format $\int p(x)dx =\int q(y)dy$ . We can then integrate both sides of this equation:

$\int f(x)dx =\int g(y)dy$

Since everything on the left is in term of $x$ and everything on the right is in terms of $y$ , we have separated the variables. We integrate the left with respect to $x$ and the right with respect to $y$ . Consider again the differential equation that describes radioactive decay: $\frac{dN}{dt}=-kN$ where $k$ is the constant of proportionality. We can separate the variables in this equation: $\frac{1}{N}dN=-kdt$ . It follows that we can integrate

$\int \frac{1}{N}dN=\int -kdt\hspace{20pt}\Longrightarrow\hspace{20pt}\ln (N)=-kt+c\hspace{20pt}\Longrightarrow\hspace{20pt}N=Ae^{-kt}$

Note that in the final step we have written $e^{-kt+c}=e^{-kt}e^c=Ae^{-kt}$ , where $A=e^c$ is also an arbitrary constant. If additional information is given in the question, it is usually possible to find the value of the proportionality and arbitrary constants. You may notice from the result above that the number of particles in radioactive decay decrease exponentially with time.

A question that requires separation of variable may expect you to draw on the integration techniques we have learnt at this level. This includes trigonometric identities, integration by substitution, integration by parts, integration by partial fractions, etc. See Example 3.

Examples

A variable $P$ grows at a rate which is inversely proportional to the square of the current time. Given that $P=5$ when $t=1$ and $P=3$ when $t=2$ , find an expression for $P$ as a function of time.

Solution:

The rate at which $P$ changes is denoted $\frac{dP}{dt}$ . This rate is inversely proportional to $t^2$ and so we can write the differential equation $\frac{dP}{dt}=\frac{k}{t^2}$ . This differential equation is simple to integrate without a separation of variables by first writing $\frac{dP}{dt}=-kt^{-2}$ . It follows that $P=kt^{-1}+c=\frac{k}{t}+c$ . When $t=1$ , $P$ must be equal to 5 and so $k+c=5$ . Also when $t=2$ , $P$ must be equal to 3 and so $\frac{k}{2}+c=3$ . Solving these simultaneous equations gives $c=1$ and $k=4$ . It follows that $P(t)=\frac{4}{t}+1$ .

The volume of a sphere is decreasing at a constant rate of $0.25{\text{cm}}^3$ every minute. What is the rate of change of the radius of the sphere when the radius is $0.25{\text{cm}}$ ?

Solution:

The volume and radius rates are related via the equation $\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}$ . We are given that $\frac{dV}{dt}=-0.25$ . We also know that the volume of the sphere is given by $V=\frac{4}{3}\pi r^3$ . Hence, $\frac{dV}{dr}=4\pi r^2$ . It follows that $\frac{dr}{dt}=\frac{dV}{dt}\div\frac{dr}{dt}=\frac{-0.25}{4\pi r^2}=-\frac{1}{16\pi r^2}$ . When $r=0.25$ , this rate is $\frac{dt}{dt}=\frac{-1}{pi}$ . At this instant, the radius is decreasing at a rate of $\frac{1}{\pi}$ cm per minute.

Find the particular solution to the differential equation $(1+x^2)\frac{dy}{dx}=2x\sec(y)$ given that $y=\frac{\pi}{2}$ when $x=0$ .

Solution:

Recall that $\sec(y)=\frac{1}{\cos(y)}$ and so we first separate the variables and integrate:

$\int\cos(y)dy=\int\frac{2x}{1+x^2}dx$

We know that the integral of $\cos(y)$ is $\sin(y)$ and we can integrate the right hand side using the substitution $u=1+x^2$ . (See Integration by Substitution). It follows that $\sin(y)=\ln\left(1+x^2\right)+c$ . Substituting $x=0$ and $y=\frac{\pi}{2}$ gives $\sin\left(\frac{\pi}{2}\right)=\ln(1)+c$ and so $c=1$ . The particular solution is $\sin(y)=\ln\left(1+x^2\right)+1$ or $y=\arcsin\left(\ln\left(1+x^2\right)+1\right)$ .