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MathidiaMathematics Academy
Differential equations
Lesson 40 of 75
Review

Differential-equation modelling

Objective. Interpret parameters, equilibrium and long-term behaviour.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

Differential-equation modelling interprets the parameters of a differential equation in context, identifying equilibrium solutions (where dy/dx=0) and determining long-term behaviour as x→∞ from the structure of the general solution. For a model like dP/dt = k(L−P), representing restricted growth towards a limit L, the equilibrium is P=L, and since the solution involves a decaying exponential, P approaches L as t increases regardless of the starting population. Interpreting the sign and size of constants, such as a decay rate or carrying capacity, is as important as solving the equation itself.

Worked example

A population P satisfies dP/dt = 0.1(500 − P), with P=100 at t=0. Find the equilibrium population and the particular solution.

A population P satisfies …
Illustrated card showing A population P satisfies dP/dt = 0
  1. 1Equilibrium occurs when dP/dt=0, giving P=500
  2. 2Separate variables: 1/(500−P) dP = 0.1 dt, integrating gives −ln|500−P| = 0.1t + c
  3. 3Rearranged: 500−P = Ae^(−0.1t); at t=0, P=100 gives 500−100=400=A
  4. 4So P = 500 − 400e^(−0.1t)

Answer: Equilibrium P=500; P = 500 − 400e^(−0.1t)

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Common mistake. Students solve the equation correctly but fail to interpret what the constant of integration or coefficient means physically, e.g. not identifying that a negative exponent causes decay to equilibrium. Always re-read the model in context after solving, checking initial conditions and long-term limits.

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