Maths with Lemon

Separable & Homogeneous Differential Equations

IB Mathematics: Analysis and Approaches Higher Level · AHL 5.18 · Lesson 10.3. Solve first-order differential equations using separation of variables and the substitution \(y=vx\), and model growth and decay.

Lesson 10.3 worksheet (PDF)Previous lesson: Euler’s MethodCalculus overview

1. Separation of Variables

What you need to know

  • A separable equation can be written as a product of a function of \(x\) and a function of \(y\).
  • Move the \(y\)-terms beside \(dy\), and the \(x\)-terms beside \(dx\).
  • Integrate both sides, including a constant.
  • Check constant solutions before dividing by \(g(y)\).

Essential formula

\[\frac{dy}{dx}=f(x)g(y)\]
\[\frac{1}{g(y)}\,dy=f(x)\,dx\]
\[\boxed{\int\frac{1}{g(y)}\,dy=\int f(x)\,dx+C}\]
Separating variables is valid where the divided expression is nonzero. Check any excluded constant solutions in the original equation.

2. Worked Example: Separable Equation

Method

  • Separate the variables.
  • Integrate both sides.
  • Rearrange the result.
  • Apply initial conditions, if given.

From the worksheet

Solve \(\frac{dy}{dx}=xy\), given \(y>0\)

Separate:

\[\frac{1}{y}\,dy=x\,dx\]

Integrate:

\[\ln y=\frac{x^2}{2}+C\]

Exponentiate:

\[\boxed{y=Ae^{x^2/2}},\quad A>0\]
You try: \(dy/dx=2y\), \(y>0\)
\[y=Ce^{2x},\quad C>0\]

3. Particular Solutions and Initial Conditions

Remember

  • A general solution includes an arbitrary constant.
  • An initial condition determines the constant.
  • Pay attention to the correct square-root branch and domain.
  • An implicit answer can sometimes be acceptable.

Worksheet example

Solve \(\frac{dy}{dx}=\frac{\sin x}{y}\), with \(y(0)=10\)
\[y\,dy=\sin x\,dx\]
\[\frac{y^2}{2}=-\cos x+C\]

Use \(y(0)=10\), giving \(C=51\).

\[\boxed{y=\sqrt{102-2\cos x}}\]

The positive branch is required by the initial condition.

4. Homogeneous Differential Equations

Recognising the structure

  • The right-hand side is a function of the ratio \(y/x\).
  • Set \(y=vx\), with \(v\) a function of \(x\).
  • Use the product rule when differentiating \(vx\).
  • The new equation is separable.

Substitution

\[\frac{dy}{dx}=F\left(\frac{y}{x}\right)\]
\[\boxed{y=vx\quad\Longrightarrow\quad\frac{dy}{dx}=v+x\frac{dv}{dx}}\]
Do not forget the extra \(v\) term from the product rule.

5. Worked Example: Homogeneous Equation

Four-step process

  • Substitute \(y=vx\).
  • Simplify to find \(dv/dx\).
  • Separate and integrate.
  • Replace \(v\) with \(y/x\).

From the worksheet

Solve \(\frac{dy}{dx}=\frac{y+x}{x}\), for \(x>0\)

Let \(y=vx\):

\[v+x\frac{dv}{dx}=v+1\]
\[x\frac{dv}{dx}=1\]
\[v=\ln x+C\]

Replace \(v=y/x\):

\[\boxed{y=x(\ln x+C)},\quad x>0\]

6. Growth, Decay and Logistic Equations

Applications

  • Radioactive decay: rate proportional to mass remaining.
  • Bacterial growth: rate proportional to population size.
  • Logistic growth: growth slows near carrying capacity.
  • The logistic equation can require partial fractions.

Key models

\[\frac{dN}{dt}=kN\quad\Longrightarrow\quad N=N_0e^{kt}\]
\[\frac{dN}{dt}=kN(a-N)\]
Radioactive decay from the worksheet

Initially 25 g, decreasing at 5 g/s. For \(dm/dt=-km\), we have \(k=1/5\).

\[\boxed{m(t)=25e^{-t/5}}\]

Half-life:

\[\boxed{t=5\ln 2\approx3.47\ \mathrm{s}}\]

7. Interactive Growth and Decay Explorer

Explore the model

  • Choose growth or decay.
  • Change the initial value and rate constant.
  • Observe how the solution changes over time.
  • Compare the exact exponential solution with your calculations.

Exact solution of \(dN/dt=kN\)

This graph uses the exact exponential solution, not Euler’s approximation.

8. HL Practice and Revision

Practice from Lesson 10.3

  • Separable equations such as \(dy/dx=x^3y^3\).
  • Homogeneous equations such as \(dy/dx=(x^2-y^2)/(xy)\).
  • Particular solutions with initial conditions.
  • Growth, decay and substitution challenges.

Common mistakes

Why check for constant solutions?

Dividing by a function of \(y\) can discard solutions where that function is zero.

What is the derivative of \(y=vx\)?
\[\frac{dy}{dx}=v+x\frac{dv}{dx}\]
When do you use partial fractions?

For integrals arising from equations such as the logistic model.

Open the complete worksheet

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