7.A Modeling Change

Activities

Building Differential Equations

Each scenario describes a system that changes over time. Write a differential equation that models the change.

  1. A population \(P\) grows at a rate proportional to the size of the population.
  2. Dead leaves \(L\) accumulate on the ground at a rate of 3 grams per square centimeter per year. At the same time, the leaves decompose at a continuous rate of 75% per year.
  3. The principal \(P\) of a home mortgage increases at 5% a year and the annual payment is $10000.
  4. Bacteria grow proportionally to the product of the current population \(N\) and the difference between the carrying capacity \(M\) and the current population \(N\).

Solve the Differential Equation

Use your knowledge of anti-derivatives to solve the following differential equations

  1. \(\displaystyle{ \frac{dy}{dx} = \cos(2 \pi x)}\)
  2. \(\displaystyle{ \frac{ds}{dt} = a t + v}\) where \(a\) and \(v\) are constants.
  3. \(\displaystyle{ \frac{dP}{dq} = \frac{100}{q}}\) where \(P(1) = 200\).
  4. \(\displaystyle{ \frac{dy}{dx} = e^{-x}}\) where \(P(2) = 0\).

Verifying Solutions to Differential Equations

Determine which of the following functions are solutions to the differential equation \[ \frac{dy}{dx} =3y. \]

  1. \(y=x^3\)
  2. \(y=e^{3x}\)
  3. \(y=\sin(3x)+\cos(3x)\)
  4. \(y=0\)

Solutions

Building Differential Equations

  1. \(\frac{dP}{dt} = r P\)
  2. \(\frac{dL}{dt} = 3 - 0.75L\)
  3. \(\frac{dP}{dt} = 0.05P - 10000\)
  4. \(\frac{dP}{dt} = k N (M-N)\)

Solve the Differential Equation

Use your knowledge of anti-derivatives to solve the following differential equations

  1. \(\displaystyle{ y = \frac{1}{2\pi} \sin(2\pi x) + C}\)
  2. \(\displaystyle{ s = \frac{1}{2}a t^2 + vt + C}\)
  3. \(\displaystyle{ P = 100 \ln (q) + 200}\)
  4. \(\displaystyle{ y = -e^{-x} + e^{-2}}\).

Verifying Solutions to Differential Equations

We must check whether \(\frac{dy}{dx} =3y.\)

  1. \(\frac{dy}{dx}=3x^2\) and \(3y=3x^3\). These are not equal, so \(y=x^3\) is NOT a solution.
  2. \(\frac{dy}{dx}=3e^{3x}\) and \(3y = 3e^{3x}\). These are equal, so \(y=e^{3x}\) is a solution!
  3. \(\frac{dy}{dx}=3\cos(3x)-3\sin(3x)\) and \(3y=3\sin(3x)+3\cos(3x)\). These are not equal (they differ by a negative sign), so \(y=\sin(3x)+\cos(3x)\) is NOT a solution.
  4. \(\frac{dy}{dx}=0\) and \(3y=0\). These are equal, so \(y=0\) is a solution!