Modelling using First Order Differential Equations (Edexcel A-Level Further Mathematics): Revision Notes

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8.1.3 Modelling using First Order Differential Equations

Introduction to Modelling with Differential Equations

First-order differential equations are used extensively to model real-world phenomena in contexts such as kinematics, population growth, and chemical reactions. A differential equation models the relationship between a rate of change and the variables involved.

Common Scenarios for Modelling

Kinematics

In kinematics, differential equations often relate velocity, acceleration, and displacement.

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Example Relationship: If v=dsdtv = \frac{ds}{dt} and a=dvdta = \frac{dv}{dt}, we can express acceleration as:

a=vdvdsa = v \frac{dv}{ds}

Population Growth or Decay

Population growth can be modelled using the equation:

dPdt=kP\frac{dP}{dt} = kP

where kk is the growth rate. This results in exponential growth or decay depending on the sign of kk.

Newton's Law of Cooling

The rate of change of temperature TT of an object is proportional to the difference between its temperature and the ambient temperature TaT_a:

dTdt=−k(T−Ta)\frac{dT}{dt} = -k(T - T_a)

Motion with Resistance

The motion of an object experiencing resistance proportional to velocity can be modelled as:

mdvdt=−kvm \frac{dv}{dt} = -kv

where kk is the resistance constant.

Worked Examples

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Example 1**: Modelling Kinematics with Resistance**

An object of mass mm falls under gravity with air resistance proportional to its velocity vv. Model and solve for v(t)v(t).

Step 1: Write the Equation

The forces acting are gravity (mgmg) and air resistance (kvkv):

mdvdt=mg−kvm \frac{dv}{dt} = mg - kv

Rearrange:

dvdt=g−kmv\frac{dv}{dt} = g - \frac{k}{m}v

Step 2: Solve the Differential Equation

This is a linear first-order differential equation:

dvdt+kmv=g\frac{dv}{dt} + \frac{k}{m}v = g

Find the integrating factor:

F(t)=e∫km dt=ekmtF(t) = e^{\int \frac{k}{m} \, dt} = e^{\frac{k}{m}t}

Multiply through by F(t)F(t):

ekmtdvdt+kmekmtv=gekmte^{\frac{k}{m}t} \frac{dv}{dt} + \frac{k}{m}e^{\frac{k}{m}t}v = ge^{\frac{k}{m}t}

Recognise as a product rule:

ddt(ekmtv)=gekmt\frac{d}{dt}\left(e^{\frac{k}{m}t}v\right) = ge^{\frac{k}{m}t}

Integrate both sides:

ekmtv=∫gekmt dt=mgkekmt+Ce^{\frac{k}{m}t}v = \int ge^{\frac{k}{m}t} \, dt = \frac{mg}{k}e^{\frac{k}{m}t} + C

Solve for vv:

v=mgk+Ce−kmtv = \frac{mg}{k} + Ce^{-\frac{k}{m}t}

Step 3: Interpret the Solution

As t→∞t \to \infty, Ce−kmt→0Ce^{-\frac{k}{m}t} \to 0, so v→mgkv \to \frac{mg}{k}.

This is the terminal velocity.

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Example 2**: Population Growth**

Model and solve a population growth scenario where the growth rate is proportional to the population, starting with P(0)=P0P(0) = P_0

Step 1: Write the Equation

dPdt=kP\frac{dP}{dt} = kP

Step 2: Solve the Differential Equation

This is separable:

1P dP=k dt\frac{1}{P} \, dP = k \, dt

Integrate both sides:

ln⁡∣P∣=kt+C\ln |P| = kt + C

Exponentiate:

P=ekt+C=CektP = e^{kt+C} = Ce^{kt}

Use the initial condition P(0)=P0P(0) = P_0

P0=Ce0  ⟹  C=P0P_0 = Ce^0 \implies C = P_0

Solution:

P(t)=P0ektP(t) = P_0 e^{kt}

Step 3: Interpret the Solution

If k>0k>0 the population grows exponentially.

If k<0k < 0 the population decays exponentially.

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Example 3**: Newton's Law of Cooling**

A hot object cools in a room at temperature Ta=20∘CT_a = 20^\circ C

Initially, the object's temperature is T(0)=100∘CT(0) = 100^\circ C

Solve for T(t)T(t)

Step 1: Write the Equation

dTdt=−k(T−Ta)\frac{dT}{dt} = -k(T - T_a)

Substitute Ta=20T_a = 20

dTdt=−k(T−20)\frac{dT}{dt} = -k(T - 20)

Step 2: Solve the Differential Equation

This is separable:

1T−20 dT=−k dt\frac{1}{T - 20} \, dT = -k \, dt

Integrate:

ln⁡∣T−20∣=−kt+C\ln|T - 20| = -kt + C

Exponentiate:

T−20=Ce−ktT - 20 = Ce^{-kt}

Solve for TT

T=20+Ce−ktT = 20 + Ce^{-kt}

Step 3: Apply the Initial Condition

When t=0,T(0)=100t = 0, T(0) = 100

100=20+C  ⟹  C=80100 = 20 + C \implies C = 80

Solution:

T(t)=20+80e−ktT(t) = 20 + 80e^{-kt}

Note Summary

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Common Mistakes:

  1. Incorrectly separating variables: Ensure terms involving tt and yy are properly isolated before integrating.
  2. Forgetting the constant of integration (CC): Always include CC after integration.
  3. Mismanaging initial conditions: Check that initial conditions are used correctly to determine constants.
  4. Incorrect signs in exponential models: Be cautious with negative exponents, especially in cooling or decay problems.
  5. Mishandling proportional relationships: Ensure constants like kk have correct interpretations (e.g., growth or decay rates).
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Key Formulas:

  1. Population Growth/Decay:
dPdt=kP  ⟹  P(t)=P0ekt\frac{dP}{dt} = kP \quad \implies \quad P(t) = P_0 e^{kt}
  1. Newton's Law of Cooling:
dTdt=−k(T−Ta)  ⟹  T(t)=Ta+Ce−kt\frac{dT}{dt} = -k(T - T_a) \quad \implies \quad T(t) = T_a + Ce^{-kt}
  1. Motion Under Gravity with Resistance:
mdvdt=mg−kv  ⟹  v(t)=mgk+Ce−kmtm \frac{dv}{dt} = mg - kv \quad \implies \quad v(t) = \frac{mg}{k} + Ce^{-\frac{k}{m}t}
  1. Linear Differential Equation (General Form):
dydx+P(x)y=Q(x)  ⟹  y=1F(x)[∫F(x)Q(x) dx+C]\frac{dy}{dx} + P(x)y = Q(x) \quad \implies \quad y = \frac{1}{F(x)} \left[ \int F(x)Q(x) \, dx + C \right]

where F(x)=e∫P(x) dxF(x) = e^{\int P(x) \, dx} is the integrating factor.

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