Setting up and integrating Problem set A Worked solutions

Write a differential equation for each situation. Define your variables first.

1. (a) The mass of mould on a wall grows at a rate proportional to its mass.

   • \(M\) = mass of mould in g, \(t\) = time in days
   • Proportional to M: \(\dfrac{dM}{dt}=kM\)

2. (b) The height of a bamboo shoot increases at a rate proportional to the cube root of its height.

   • \(h\) = height in cm, \(t\) = time in days
   • Cube root is the power 1/3: \(\dfrac{dh}{dt}=kh^{1/3}\)

Write a differential equation for each situation. Define your variables first.

1. (a) The thickness of ice on a pond increases at a rate inversely proportional to the thickness.

   • \(x\) = thickness in cm, \(t\) = time in hours
   • Inverse proportion: \(\dfrac{dx}{dt}=\dfrac kx\)

2. (b) The radius of a growing bubble increases at a rate inversely proportional to the cube of the radius.

   • \(r\) = radius in mm, \(t\) = time in s
   • Inverse to the cube: \(\dfrac{dr}{dt}=\dfrac{k}{r^3}\)

Write a differential equation for each situation. Define your variables first.

1. (a) The number of unsold tickets decreases at a rate proportional to the square root of that number.

   • \(N\) = unsold tickets, \(t\) = time in days
   • A decrease: \(\dfrac{dN}{dt}=-k\sqrt N\) with \(k>0\)

2. (b) The speed of a crate sliding down a ramp decreases at a rate inversely proportional to the square of its speed.

   • \(v\) = speed in m s⁻¹, \(t\) = time in s
   • Inverse to the square, and a decrease: \(\dfrac{dv}{dt}=-\dfrac{k}{v^2}\)

Find the general solution of each differential equation.

1. (a) \(\dfrac{dy}{dx}=2x^{3} + \frac{6}{\sqrt{x}}\)

   • Integrate both sides with respect to \(x\)
   • \(y=12 \sqrt{x} + \frac{x^{4}}{2}+c\)

2. (b) \(\dfrac{dy}{dt}=5e^{-\frac{t}{2}}\)

   • Integrate both sides with respect to \(t\)
   • \(y=-10e^{-\frac{t}{2}}+c\)

3. (c) \(\dfrac{dy}{dt}=2 + \frac{8}{t}\), \(t>0\)

   • Integrate both sides with respect to \(t\)
   • \(y=2t + 8 \ln|t|+c\)

4. (d) \(\dfrac{ds}{dt}=4 \cos(2t)\)

   • Integrate both sides with respect to \(t\)
   • \(s=2 \sin(2t)+c\)

A town has 5000 residents. After \(t\) hours, \(S\) of them have seen a viral clip. The rate at which \(S\) increases is proportional to the number who have seen the clip and to the number who have not. Write a differential equation.

• \(S\) = number who have seen the clip, \(t\) = hours
• The number who have not seen it is \(5000-S\)
• \(\dfrac{dS}{dt}=kS(5000-S)\)

A hashtag spreads through a group of 3000 people. After \(t\) hours, \(H\) of them have used it. The rate at which \(H\) increases is proportional to the number who have used it and to the number who have not, and is inversely proportional to \(t\). Write a differential equation.

• \(H\) = number who have used the hashtag, \(t\) = hours
• Not used it: \(3000-H\)
• Multiply, then divide by \(t\): \(\dfrac{dH}{dt}=\dfrac{kH(3000-H)}{t}\)

A cyclist's velocity is \(v=\dfrac{ds}{dt}=6e^{-0.2t}\) m s⁻¹, where \(s\) is the distance from the start in metres and \(t\) is the time in seconds.

1. (a) Find the general solution for \(s\) in terms of \(t\).

   • Integrate with respect to \(t\)
   • \(s=-30e^{-\frac{t}{5}}+c\)

2. (b) Given that the cyclist starts at \(s=0\), find the distance after 10 seconds.

   • At \(t=0\), \(s=0\): \(c=30\)
   • At \(t=10\): \(s=30 -\frac{30}{e^{2}}=25.9\)
   • 25.9 m

Rearrange each equation to make the derivative the subject, then find the general solution.

1. (a) \(\dfrac{dF}{dm}+\frac{2}{m^{2}}=5\), \(m>0\)

   • \(\dfrac{dF}{dm}=5-\frac{2}{m^{2}}\)
   • Integrate both sides: \(F=5m + \frac{2}{m}+c\)

2. (b) \(\dfrac{dG}{dt}+8e^{-2t}=1\)

   • \(\dfrac{dG}{dt}=1-8e^{-2t}\)
   • Integrate both sides: \(G=t + 4e^{-2t}+c\)