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AI HL · AHL 9

Differential Equations

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AHL 9.1

Setting up differential equations and integrating dy/dx = f(x)

AHL 9.1

Setting up differential equations and integrating dy/dx = f(x)

Target: I can define my variables and write a rate of change as a differential equation with a constant of proportionality k. (+4 more)

By the end of this section

Key points

Writing the equation

  • Say what each letter means first, with units: \(V\) = volume left in litres, \(t\) = time in minutes. Then write the rate as \(\dfrac{dV}{dt}\).
  • "Proportional to" means multiplied by a constant: \(\dfrac{dV}{dt}=kV\). "Inversely proportional to" puts the quantity underneath: \(\dfrac{dV}{dt}=\dfrac{k}{V}\).
  • "Square root of V" is \(V^{1/2}\), "cube root" is \(V^{1/3}\). "Square of V" is \(V^2\).
  • A decrease needs a minus sign, written in front of \(k\), and \(k\) stays positive: \(\dfrac{dV}{dt}=-k\sqrt V\).

Two quantities at once

  • If the rate is proportional to two quantities, multiply them. In a group of \(N\) people with \(I\) infected, the number not infected is \(N-I\), so \(\dfrac{dI}{dt}=kI(N-I)\).
  • "Also inversely proportional to the time \(t\)" means divide the whole right-hand side by \(t\).

Integrating dy/dx = f(x)

  • Integrate both sides with respect to \(x\). The answer is a general solution: it has a constant \(c\) and stands for a whole family of curves.
  • Useful results: \(\int e^{kt}dt=\dfrac{e^{kt}}{k}+c\), \(\int\dfrac1t\,dt=\ln|t|+c\), \(\int\sin kt\,dt=-\dfrac{\cos kt}{k}+c\), \(\int\cos kt\,dt=\dfrac{\sin kt}{k}+c\).
  • Do not forget the \(+c\). It is the first thing a marker looks for.

Rearranging first

  • In \(\dfrac{dF}{dm}+g(m)=c\) the derivative is not alone. Move \(g(m)\) across to get \(\dfrac{dF}{dm}=c-g(m)\), then integrate.
  • The sign changes when a term moves across. Check by differentiating your answer.

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
A decrease proportional to a root

The area of a patch of snow decreases at a rate proportional to the fourth root of its area. Write a differential equation for this. What power of the area appears on the right-hand side?

Working
  1. Define the variables

    \(A\) = area of snow in m², \(t\) = time in hours

    Say what the letters mean, with units.

  2. Write the rate

    \(\dfrac{dA}{dt}\)

    The rate of change of the area.

  3. Proportional to a root

    \(\dfrac{dA}{dt}=kA^{1/4}\)

    Proportional means multiplied by a constant k. A fourth root is the power 1/4.

  4. A decrease

    \(\dfrac{dA}{dt}=-kA^{1/4},\ k>0\)

    The area is falling, so the rate is negative. The minus goes in front of k.

Video walkthrough coming
An inverse proportion

The speed \(v\) of a sled sliding on rough snow decreases at a rate inversely proportional to the square of the speed. Write a differential equation. Then find the right-hand side when \(v = 2\) and \(k = 8\).

Working
  1. Define the variables

    \(v\) = speed in m s⁻¹, \(t\) = time in s

    Units first.

  2. Inverse proportion

    \(\dfrac{dv}{dt}=\dfrac{k}{v^2}\)

    Inversely proportional to v² puts v² in the denominator.

  3. Decrease

    \(\dfrac{dv}{dt}=-\dfrac{k}{v^2},\ k>0\)

    The speed falls, so add the minus sign.

  4. Put in the values

    \(-\dfrac{8}{2^2}=-2\)

    The speed is falling at 2 m s⁻² at that moment.

Video walkthrough coming
Proportional to two quantities

In a school of \(N\) students, \(S\) have signed a petition. The rate at which \(S\) increases is proportional to the number who have signed and to the number who have not. Write a differential equation. Find the rate when \(N = 100\), \(S = 30\) and \(k = 0.002\). Then say how the equation changes if the rate is also inversely proportional to the time \(t\).

Working
  1. Variables

    \(S\) = number who have signed, \(t\) = days since the petition began

    Define each letter.

  2. Who has not signed

    \(N-S\)

    The number who have not signed.

  3. Multiply the factors

    \(\dfrac{dS}{dt}=kS(N-S)\)

    Proportional to both: multiply them together with one constant k.

  4. Put in the values

    \(0.002\times30\times70=4.2\)

    N − S = 100 − 30 = 70.

  5. Also inversely proportional to time

    \(\dfrac{dS}{dt}=\dfrac{kS(N-S)}{t}\)

    Divide the whole right-hand side by t.

Video walkthrough coming
General solution by integrating

Given that \(\dfrac{dh}{dt}=6e^{-3t}\), find \(h\) in terms of \(t\). Give the number that multiplies \(e^{-3t}\).

Working
  1. Integrate both sides

    \(h=\int6e^{-3t}\,dt\)

    Undo the differentiation.

  2. Divide by the coefficient of t

    \(h=\dfrac{6}{-3}e^{-3t}+c\)

    The integral of e^(kt) is e^(kt)/k. Here k = −3.

  3. General solution

    \(h=-2e^{-3t}+c\)

    Keep the constant c. Check: differentiating gives 6e^(−3t).

Video walkthrough coming
Rearrange, then integrate

Find the general solution of \(\dfrac{dF}{dm}+\dfrac{4}{m}=2\) for \(m>0\). Give the number that multiplies \(\ln m\).

Working
  1. Get the derivative alone

    \(\dfrac{dF}{dm}=2-\dfrac4m\)

    Subtract 4/m from both sides.

  2. Integrate

    \(F=\int\left(2-\dfrac4m\right)dm\)

    Integrate each term.

  3. General solution

    \(F=2m-4\ln m+c\)

    m > 0, so no modulus is needed.

Video walkthrough coming