Constructing differential equationsAQA A-Level Maths: Revision notes

A differential equation links a quantity to its rate of change. Look for the quantity, the variable it changes with, and a phrase describing the rate.

• 'rate of increase of $N$ with time' means $\frac{dN}{dt}$
• '$y$ is proportional to $x$' means $y=kx$; so 'the rate is proportional to $N$' means $\frac{dN}{dt}=kN$
• 'proportional to the square root of $V$' means $\frac{dV}{dt}=\pm k\sqrt V$
• 'inversely proportional' means a quotient, such as $\frac{dN}{dt}=\frac kN$ Decrease means a negative rate. Write the minus sign in the equation and keep the constant $k$ positive. Here you only construct the equation; you are not asked to solve it.

If the rate of growth of a population $N$ is proportional to $N$: $\frac{dN}{dt}=kN.$ Given a pair of values, find $k$: with $N=2000$ and $\frac{dN}{dt}=300$, $k=0.15$. Combine processes by adding rates: net rate = rate in $-$ rate out. Growth proportional to $N$ with a constant removal of $40$ per hour gives $\frac{dN}{dt}=0.15N-40.$ The population is steady when $\frac{dN}{dt}=0$. Here that needs $N=\frac{40}{0.15}=266.7$.

Velocity and acceleration are rates of change: $v=\frac{ds}{dt}$ and $a=\frac{dv}{dt}$. Newton's second law $F=ma$ gives $m\frac{dv}{dt}=$ resultant force. A parachutist falling with weight $mg$ and air resistance $\lambda v$: $m\frac{dv}{dt}=mg-\lambda v\quad\Rightarrow\quad\frac{dv}{dt}=g-\frac{\lambda}{m}v.$ If the resistance is $\mu v^2$, then $\frac{dv}{dt}=g-\frac{\mu}{m}v^2$. Terminal speed is reached when the acceleration is zero: $v=\frac{mg}{\lambda}$ in the first model and $v=\sqrt{\frac{mg}{\mu}}$ in the second.

Demand $Q$ depends on price $p$, so the relevant rate is $\frac{dQ}{dp}$ with respect to price, not time. If $Q$ falls as $p$ rises and the fall is proportional to $Q$: $\frac{dQ}{dp}=-kQ,\qquad k>0.$ Given $Q=40$ and $\frac{dQ}{dp}=-6$ at $p=5$: $k=0.15$. Revenue is $R=pQ$. By the product rule, $\frac{dR}{dp}=Q+p\frac{dQ}{dp}=Q(1-kp)$. Revenue is stationary when $1-kp=0$, so at $p=\frac1k=\pounds6.67$.

For flow problems, write net rate = inflow $-$ outflow. A tank that drains at a rate proportional to $\sqrt V$ and fills at $0.3$ m$^3$ per minute has $\frac{dV}{dt}=0.3-k\sqrt V.$ With $V=16$ and a loss of $0.8$ per minute, $k\sqrt{16}=0.8$, so $k=0.2$. The volume is constant when $0.3=0.2\sqrt V$, so $V=2.25$ m$^3$. Always check units, check that the signs agree with the story, and say what the constant means.