Separable equations in context: growth, decay and harder forms Problem set A Worked solutions

A sample of a radioactive substance decays at a rate proportional to its mass \(m\) g. At first the mass is 60 g and it is decreasing at 7.2 g per day.

1. (a) Find the constant \(k\) in \(\dfrac{dm}{dt}=-km\).

   • \(-7.2=-k\times60\)
   • \(k=0.12\)

2. (b) Find \(m\) in terms of \(t\).

   • Separate: \(\int\dfrac1m\,dm=\int-0.12\,dt\)
   • \(\ln m=-0.12t+c\)
   • At \(t=0\), \(m=60\): \(m=60e^{-0.12t}\)

3. (c) Find how long it takes for the mass to halve, to 3 significant figures.

   • \(\tfrac12\times60=60e^{-0.12t}\)
   • \(t=\dfrac{\ln2}{0.12}=5.78\)

The number \(N\) of yeast cells in a vat grows at a rate proportional to \(N\). At first there are 800 cells and the number is increasing at 56 cells per hour.

1. (a) Find \(k\) in \(\dfrac{dN}{dt}=kN\).

   • \(56=k\times800\)
   • \(k=0.07\)

2. (b) Find the number after 6 hours, to 3 significant figures.

   • \(N=800e^{0.07t}\)
   • At \(t=6\): \(N=800e^{0.42}=1220\)

Solve each differential equation.

1. (a) Find the general solution of \(\dfrac{dy}{dx}=e^{3x+y}\).

   • \(\dfrac{dy}{dx}=e^{3x}e^{y}\)
   • \(\int e^{-y}\,dy=\int e^{3x}\,dx\)
   • \(-e^{-y}=\tfrac13e^{3x}+c\)
   • \(e^{-y}=A-\tfrac13e^{3x},\ \ y=-\ln\left(A-\tfrac13e^{3x}\right)\)

2. (b) Find the particular solution of \(\dfrac{dy}{dx}=5e^{x-4y}\), \(y=0\) when \(x=0\). Give \(y\) in terms of \(x\).

   • \(\int e^{4y}\,dy=\int5e^{x}\,dx\)
   • \(\tfrac14e^{4y}=5e^{x}+c\)
   • At \(x=0\), \(y=0\): \(\tfrac14=5+c\Rightarrow c=-\tfrac{19}{4}\)
   • \(e^{4y}=20e^{x}-19,\ \ y=\tfrac14\ln\left(20e^{x}-19\right)\)

Given that \(y\dfrac{dy}{dx}=3e^{-x}\) and that \(y=-2\) when \(x=0\), find \(y\) in terms of \(x\).

• \(\int y\,dy=\int3e^{-x}\,dx\)
• \(\dfrac{y^2}{2}=-3e^{-x}+c\)
• \(x=0,\ y=-2:\ 2=-3+c\Rightarrow c=5\)
• \(y^2=10-6e^{-x}\)
• \(y\) was negative at the start: \(y=-\sqrt{10-6e^{-x}}\)

Water flows into a tank so that the depth \(h\) cm increases at a rate inversely proportional to \(h\). When \(h=40\) the depth is increasing at 0.5 cm per minute. \(t\) is the time in minutes.

1. (a) Show that \(\dfrac{dh}{dt}=\dfrac{20}{h}\).

   • \(\dfrac{dh}{dt}=\dfrac{k}{h}\)
   • \(0.5=\dfrac{k}{40}\Rightarrow k=20\)
   • So \(\dfrac{dh}{dt}=\dfrac{20}{h}\)

2. (b) Find \(h\) in terms of \(t\).

   • \(\int h\,dh=\int 20\,dt\)
   • \(\dfrac{h^2}{2}=20t+c\)
   • At \(t=0\): \(c=800\)
   • \(h^2=40t+1600\)

3. (c) Find the time taken for the depth to reach 60 cm, to 3 significant figures.

   • \(3600=40t+1600\)
   • \(t=50.0\)

An object falls through air. Its velocity \(v\) m s⁻¹ after \(t\) seconds satisfies \(\dfrac{dv}{dt}=14-0.35v\), and it starts from rest.

1. (a) Find \(v\) in terms of \(t\).

   • \(\int\dfrac{1}{14-0.35v}\,dv=\int1\,dt\)
   • \(-\dfrac{1}{0.35}\ln(14-0.35v)=t+c\)
   • \(14-0.35v=Ae^{-0.35t},\ A=14 \text{ since } v=0 \text{ at } t=0\)
   • \(v=40\left(1-e^{-0.35t}\right)\)

2. (b) Find the distance it travels in the first 3 seconds, to 3 significant figures.

   • From the calculator: \(\int_0^{3}40\left(1-e^{-0.35t}\right)\,dt=45.7\)

A spherical mint dissolves. Its volume decreases at a rate proportional to its surface area. When the volume is 12 cm³ it is decreasing at 1.5 cm³ per minute. Let \(r\) cm be the radius and \(t\) the time in minutes.

1. (a) Find \(\dfrac{dr}{dt}\).

   • \(\dfrac{dV}{dt}=-k\times4\pi r^2=4\pi r^2\dfrac{dr}{dt}\Rightarrow\dfrac{dr}{dt}=-k\)
   • \(12=\tfrac43\pi r^3\Rightarrow r=1.42,\ S=4\pi r^2=25.35\)
   • \(1.5=k\times25.35\Rightarrow k=0.0592\)
   • \(\dfrac{dr}{dt}=-0.0592\)

2. (b) Find the time taken to dissolve completely.

   • \(t=\dfrac{1.42}{0.0592}=24.0\)

A metal block is heated to 150 °C and placed in a room at 22 °C. Its temperature \(T\) °C after \(t\) minutes decreases at a rate proportional to \(T-22\). After 6 minutes the block is at 100 °C.

1. (a) Write down a differential equation for \(T\), using a constant \(k\).

   • The rate is proportional to \(T-22\) and \(T\) is falling
   • \(\dfrac{dT}{dt}=-k(T-22)\) with \(k>0\)

2. (b) Find \(T\) in terms of \(t\).

   • \(\int\dfrac{1}{T-22}\,dT=\int-k\,dt\), so \(T=22+Be^{-kt}\)
   • \(T=150\) at \(t=0\): \(B=128\)
   • \(T=100\) at \(t=6\): \(78=128e^{-6k}\), so \(k=\dfrac{\ln(128/78)}{6}=0.0826\)
   • \(T=22+128e^{-0.0826t}\)

3. (c) Find how long it takes to cool to 40 °C, to 3 significant figures.

   • \(40=22+128e^{-kt}\)
   • \(t=\dfrac{\ln(128/18)}{k}=23.8\)

A tank holds 300 litres of water in which 10 kg of salt is dissolved. Salt solution of concentration 0.1 kg per litre flows in at 6 litres per minute. The mixture is stirred and flows out at 6 litres per minute. After \(t\) minutes the tank holds \(m\) kg of salt.

1. (a) Show that \(\dfrac{dm}{dt}=0.6-0.02m\).

   • In: \(0.1\times6=0.6\) kg per minute
   • Out: \(\dfrac{m}{300}\times6=0.02m\) kg per minute
   • \(\dfrac{dm}{dt}=0.6-0.02m\)

2. (b) Find \(m\) in terms of \(t\).

   • \(\int\dfrac{1}{0.6-0.02m}\,dm=\int dt\), so \(0.6-0.02m=Ae^{-0.02t}\)
   • \(m=10\) at \(t=0\): \(A=0.4\)
   • \(m=30-20e^{-0.02t}\)

3. (c) Find the time taken for the mass to reach 20 kg, to 3 significant figures.

   • \(20=30-20e^{-0.02t}\)
   • \(e^{-0.02t}=\dfrac{10}{20}\)
   • \(t=34.7\)

4. (d) State the mass the tank approaches in the long run.

   • As \(t\to\infty\), \(e^{-0.02t}\to0\)
   • \(m\to30\) kg (this is \(0.6\div0.02\))

Given that \(\dfrac{dy}{dx}=\dfrac{2\cos x}{\sin y}\):

1. (a) Show that the general solution can be written as \(\cos y+2\sin x=c\) for some constant \(c\).

   • \(\int\sin y\,dy=\int2\cos x\,dx\)
   • \(-\cos y=2\sin x+c_1\)
   • \(\cos y+2\sin x=c\)

2. (b) A particular solution satisfies \(y=\dfrac{2\pi}{3}\) when \(x=\dfrac\pi2\), for \(0\le x\le\pi\) and \(-\pi\le y\le\pi\). Find the two possible values of \(y\) when \(x=\dfrac{5\pi}{6}\), to 3 significant figures.

   • \(\cos\tfrac{2\pi}{3}+2\sin\tfrac\pi2=-\tfrac12+2=\tfrac32\Rightarrow c=\tfrac32\)
   • \(\cos y=\tfrac32-2\sin\tfrac{5\pi}{6}=\tfrac12\)
   • \(y=\pm\tfrac\pi3=\pm1.05\)