Solving Bernoulli-type differential equation

This post guides you through solving the Bernoulli-type differential equation \(y' = 2y - y^2\) step-by-step, using simple explanations and math reasoning. Click on any menu item to jump directly to that step.

Problem: \(y' = 2y - y^2\)

Step 1 – Identify Type Step 2 – Divide by \(y^2\) Step 3 – Substitution Step 4 – Homogeneous Part Step 5 – Variation of Constant Step 6 – Back to \(y\) Step 7 – Equilibria & Stability Extra – What Is Variation of Constant?

Step 1 – Identify the Equation Type

The equation contains both \(y\) and \(y^2\) terms, so it’s a Bernoulli equation of the form:

\(y' = P(x)y + Q(x)y^n\)

Comparing terms gives \(P(x) = 2,\ Q(x) = -1,\ n = 2.\)

Why Bernoulli? It mixes a first-power and a squared term of \(y\). This type can be transformed into a linear equation using a special substitution.

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Step 2 – Divide by \(y^2\)

Divide the entire equation by \(y^2\):

\(y^{-2}y' = 2y^{-1} - 1\)

Why? This step prepares the equation for substitution by producing \(y^{-2}y'\), which appears when differentiating \(y^{-1}\).

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Step 3 – Substitution \(z = y^{-1}\)

Let \(z = \frac{1}{y}\). Then differentiate: \(z' = -y^{-2}y'\).

Substitute this into the equation:

\(-z' = 2z - 1 \Rightarrow z' = -2z + 1\)

This is now a linear differential equation in \(z\):

\(z' + 2z = 1\)

Why it helps: The substitution removes the non-linear part (\(y^2\)) and gives a linear equation that is easier to solve.

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Step 4 – Solve the Homogeneous Part

First, solve \(z'_h + 2z_h = 0\).

\(\frac{dz}{dt} = -2z\)

Separate and integrate:

\(\int \frac{1}{z}dz = \int -2dt \Rightarrow \ln z = -2t + C\)

Exponentiate:

\(z_h = C_1 e^{-2t}\)

Meaning: This is the base part of the solution — how \(z\) behaves naturally when there’s no external term on the right-hand side.

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Step 5 – Apply Variation of Constant

Let \(C_1\) become a function of time: \(C_1 = C_1(t)\).

Write \(z = e^{-2t}C_1(t)\) and find its derivative:

\(z' = -2e^{-2t}C_1 + e^{-2t}C_1'\)

Substitute into \(z' + 2z = 1\):

\(-2e^{-2t}C_1 + e^{-2t}C_1' + 2e^{-2t}C_1 = 1\)

After cancellation:

\(e^{-2t}C_1' = 1 \Rightarrow C_1' = e^{2t}\)

Integrate both sides:

\(C_1(t) = \tfrac{1}{2}e^{2t} + C_2\)

What’s happening? We allow the “constant” to change over time so the equation can balance with the non-homogeneous term (+1).

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Step 6 – Back-Substitute to Find \(y\)

Substitute back: \(z = e^{-2t}(\tfrac{1}{2}e^{2t} + C_2)\)

\(z = \tfrac{1}{2} + C_2 e^{-2t}\)

Since \(z = 1/y\):

\(\tfrac{1}{y} = \tfrac{1}{2} + C_2 e^{-2t}\)

Take the reciprocal:

\(y = \frac{2}{1 + 2C_2 e^{-2t}} \Rightarrow y = \frac{2}{1 + C e^{-2t}}\)

Interpretation: The value of \(y\) grows quickly when small but gradually slows and approaches 2 as time increases.

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Step 7 – Equilibrium Solutions and Stability

Set \(y' = 0\):

\(y(2 - y) = 0 \Rightarrow y = 0 \text{ or } y = 2\)

• \(y=0\): small increase makes \(y'\) positive → unstable.
• \(y=2\): small changes return to 2 → stable.

Conclusion: \(y=2\) is a stable equilibrium point, and \(y=0\) is unstable.

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Extra – What Does “Variation of Constant” Mean?

In a homogeneous equation, the constant \(C\) is fixed. But when we have an added term (like +1), that constant must adapt over time. We treat it as a function \(C(t)\), find its derivative, and integrate. This trick lets us capture both the natural and forced changes of the system.

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