Constructing differential equationsEdexcel A-Level Maths: Revision notes

A differential equation involves a derivative. To construct one, identify the quantity that changes, the variable it changes with respect to, and the stated relationship. 'The rate of increase of $N$' means $\frac{dN}{dt}$.

• 'Proportional to $N$': $\frac{dN}{dt}=kN$.
• 'Inversely proportional to $r^{2}$': $\frac{dr}{dt}=\frac{k}{r^{2}}$.
• 'Decreasing' or 'rate of decrease': a negative sign, so that $k$ can stay positive. Use the stated sign of $k$ and the direction of change to decide whether the right-hand side is positive or negative.

Use the information given to find $k$ by substitution. Example: insects with $N=500$ increasing at $60$ per day: $60=k\times500$, so $k=0.12$ and $\frac{dN}{dt}=0.12N$. When $N=2000$ the rate is $240$ per day. Example: melting ice with $\frac{dr}{dt}=-\frac{k}{r^{2}}$ and $\frac{dr}{dt}=-0.1$ at $r=2$: $0.1=\frac k4$, so $k=0.4$. When $r=0.5$, $\frac{dr}{dt}=-1.6$.

When something enters and leaves a system, the net rate is the rate in minus the rate out. Example: a tank with $V=2h$, inflow $0.6$ and outflow $0.1\sqrt h$. Then $2\frac{dh}{dt}=0.6-0.1\sqrt h$, so $\frac{dh}{dt}=0.3-0.05\sqrt h$. The depth stops changing when this is zero: $\sqrt h=6$, $h=36$. Example: a drug delivered at $5$ mg per hour and removed at rate $kx$ gives $\frac{dx}{dt}=5-kx$. If $x=40$ is steady then $0=5-40k$ and $k=\frac18$. A value of the variable where the derivative is zero is an equilibrium.

The same approach works in many contexts.

• Kinematics: acceleration $\frac{dv}{dt}$ proportional to velocity, resisting motion: $\frac{dv}{dt}=-kv$.
• Population growth: $\frac{dP}{dt}=kP$ for unrestricted growth.
• Price and demand: demand $D$ falling at a rate proportional to $D$ as price $p$ rises: $\frac{dD}{dp}=-kD$. Different models can be compared by evaluating their predicted rates. If a second drug model has removal proportional to $\sqrt x$ with $x=40$ steady, $c=\frac{5}{\sqrt{40}}\approx0.791$ and at $x=16$ the rate is $1.84$, lower than $3$ from the first model. At this stage you construct the equation; solving it comes later.