Separable equations in context: growth, decay and harder forms Problem set B One question at a time

The mass \(m\) mg of caffeine in a person's blood decreases at a rate proportional to \(m\). At first \(m=120\) and it is decreasing at 15 mg per hour.

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

   • \(-15=-k\times120\)
   • \(k=0.125\)

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

   • Separate: \(\int\dfrac1m\,dm=\int-0.125\,dt\)
   • \(\ln m=-0.125t+c\)
   • At \(t=0\), \(m=120\): \(m=120e^{-0.125t}\)

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

   • \(\tfrac12\times120=120e^{-0.125t}\)
   • \(t=\dfrac{\ln2}{0.125}=5.55\)

The number \(N\) of members of an online forum grows at a rate proportional to \(N\). At first there are 2500 members and the number is increasing at 100 per day.

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

   • \(100=k\times2500\)
   • \(k=0.04\)

2. (b) Find the number after 15 days, to 3 significant figures.

   • \(N=2500e^{0.04t}\)
   • At \(t=15\): \(N=2500e^{0.6}=4560\)

Solve each differential equation.

1. (a) Find the general solution of \(\dfrac{dy}{dx}=e^{2x+3y}\).

   • \(\dfrac{dy}{dx}=e^{2x}e^{3y}\)
   • \(\int e^{-3y}\,dy=\int e^{2x}\,dx\)
   • \(-\tfrac13e^{-3y}=\tfrac12e^{2x}+c\)
   • \(e^{-3y}=A-\tfrac32e^{2x},\ \ y=-\tfrac13\ln\left(A-\tfrac32e^{2x}\right)\)

2. (b) Find the particular solution of \(\dfrac{dy}{dx}=3e^{2x-y}\), \(y=0\) when \(x=0\). Give \(y\) in terms of \(x\).

   • \(\int e^{y}\,dy=\int3e^{2x}\,dx\)
   • \(e^{y}=\tfrac32e^{2x}+c\)
   • At \(x=0\), \(y=0\): \(1=\tfrac32+c\Rightarrow c=-\tfrac12\)
   • \(y=\ln\left(\tfrac32e^{2x}-\tfrac12\right)\)

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

• \(\int y\,dy=\int2e^{-2x}\,dx\)
• \(\dfrac{y^2}{2}=-e^{-2x}+c\)
• \(x=0,\ y=-3:\ 4.5=-1+c\Rightarrow c=5.5\)
• \(y^2=11-2e^{-2x}\)
• \(y\) was negative at the start: \(y=-\sqrt{11-2e^{-2x}}\)

An oil slick spreads so that its radius \(r\) m increases at a rate inversely proportional to \(r\). When \(r=25\) the radius is increasing at 0.4 m per hour. \(t\) is the time in hours.

1. (a) Show that \(\dfrac{dr}{dt}=\dfrac{10}{r}\).

   • \(\dfrac{dr}{dt}=\dfrac{k}{r}\)
   • \(0.4=\dfrac{k}{25}\Rightarrow k=10\)
   • So \(\dfrac{dr}{dt}=\dfrac{10}{r}\)

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

   • \(\int r\,dr=\int 10\,dt\)
   • \(\dfrac{r^2}{2}=10t+c\)
   • At \(t=0\): \(c=312.5\)
   • \(r^2=20t+625\)

3. (c) Find the time taken for the radius to reach 40 m, to 3 significant figures.

   • \(1600=20t+625\)
   • \(t=48.8\)

A ball sinks through syrup. Its velocity \(v\) m s⁻¹ after \(t\) seconds satisfies \(\dfrac{dv}{dt}=9-0.45v\), and it starts from rest.

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

   • \(\int\dfrac{1}{9-0.45v}\,dv=\int1\,dt\)
   • \(-\dfrac{1}{0.45}\ln(9-0.45v)=t+c\)
   • \(9-0.45v=Ae^{-0.45t},\ A=9 \text{ since } v=0 \text{ at } t=0\)
   • \(v=20\left(1-e^{-0.45t}\right)\)

2. (b) Find the distance it travels in the first 4 seconds, to 3 significant figures.

   • From the calculator: \(\int_0^{4}20\left(1-e^{-0.45t}\right)\,dt=42.9\)

A spherical ball of ice melts. Its volume decreases at a rate proportional to its surface area. When the volume is 30 cm³ it is decreasing at 4 cm³ per hour. Let \(r\) cm be the radius and \(t\) the time in hours.

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

   • \(\dfrac{dV}{dt}=-k\times4\pi r^2=4\pi r^2\dfrac{dr}{dt}\Rightarrow\dfrac{dr}{dt}=-k\)
   • \(30=\tfrac43\pi r^3\Rightarrow r=1.928,\ S=4\pi r^2=46.69\)
   • \(4=k\times46.69\Rightarrow k=0.0857\)
   • \(\dfrac{dr}{dt}=-0.0857\)

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

   • \(t=\dfrac{1.928}{0.0857}=22.5\)

A bowl of soup at 85 °C is left in a kitchen at 18 °C. Its temperature \(T\) °C after \(t\) minutes decreases at a rate proportional to \(T-18\). After 8 minutes the soup is at 60 °C.

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

   • The rate is proportional to \(T-18\) and \(T\) is falling
   • \(\dfrac{dT}{dt}=-k(T-18)\) with \(k>0\)

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

   • \(\int\dfrac{1}{T-18}\,dT=\int-k\,dt\), so \(T=18+Be^{-kt}\)
   • \(T=85\) at \(t=0\): \(B=67\)
   • \(T=60\) at \(t=8\): \(42=67e^{-8k}\), so \(k=\dfrac{\ln(67/42)}{8}=0.0584\)
   • \(T=18+67e^{-0.0584t}\)

3. (c) Find how long it takes to cool to 30 °C, to 3 significant figures.

   • \(30=18+67e^{-kt}\)
   • \(t=\dfrac{\ln(67/12)}{k}=29.5\)

A vat holds 500 litres of syrup in which 25 kg of sugar is dissolved. Sugar solution of concentration 0.25 kg per litre flows in at 8 litres per minute. The mixture is stirred and flows out at 8 litres per minute. After \(t\) minutes the vat holds \(m\) kg of sugar.

1. (a) Show that \(\dfrac{dm}{dt}=2-0.016m\).

   • In: \(0.25\times8=2\) kg per minute
   • Out: \(\dfrac{m}{500}\times8=0.016m\) kg per minute
   • \(\dfrac{dm}{dt}=2-0.016m\)

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

   • \(\int\dfrac{1}{2-0.016m}\,dm=\int dt\), so \(2-0.016m=Ae^{-0.016t}\)
   • \(m=25\) at \(t=0\): \(A=1.6\)
   • \(m=125-100e^{-0.016t}\)

3. (c) Find the time taken for the mass to reach 100 kg, to 3 significant figures.

   • \(100=125-100e^{-0.016t}\)
   • \(e^{-0.016t}=\dfrac{25}{100}\)
   • \(t=86.6\)

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

   • As \(t\to\infty\), \(e^{-0.016t}\to0\)
   • \(m\to125\) kg (this is \(2\div0.016\))

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

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

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

2. (b) A particular solution satisfies \(y=\dfrac\pi3\) when \(x=\dfrac\pi2\), for \(0\le x\le\pi\) and \(-\pi\le y\le\pi\). Find the two possible values of \(y\) when \(x=\dfrac\pi6\), to 3 significant figures.

   • \(2\cos\tfrac\pi3+\sin\tfrac\pi2=1+1=2\Rightarrow c=2\)
   • \(\cos y=\dfrac{2-\sin\frac\pi6}{2}=0.75\)
   • \(y=\pm0.723\)