Unit 7: Differential Equations

Separable Differential Equations

Solving separable differential equations, including exponential growth and decay models.

Advanced25 min lesson3 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

Slope Fields

Solving a Differential Equation with Integration

A slope field shows what a solution looks like, but not its exact formula. Before reading on: for dy/dx = ky (a rate proportional to the current amount — exactly the exponential growth/decay setup), could separating the x's and y's onto opposite sides of the equation, then integrating both sides, actually solve for y directly?

Definition — Separable Differential Equation

A differential equation is separable when it can be rewritten with all y-terms (including dy) on one side and all x-terms (including dx) on the other. Integrating both sides separately, then solving for y, gives the general solution; a given initial condition pins down the constant of integration.

Solving dy/dx = ky by separating variables: dy/y = k dx. Integrate both sides: ∫dy/y = ∫k dx, giving ln|y| = kx + C₁. Exponentiate both sides: |y| = e^(kx + C₁) = e^(C₁)·e^(kx). Writing A = ±e^(C₁) to absorb the sign and constant: y = Ae^(kx). Applying an initial condition y(0) = y₀ gives y₀ = Ae⁰ = A, so A = y₀ — producing y = y₀e^(kx), the exponential growth/decay model, now derived directly from its differential equation rather than simply stated.

\frac{dy}{dx} = ky \;\Rightarrow\; y = y_0 e^{kx}

Worked Example — Solving a Separable Equation Step by Step

Solve dy/dx = 2xy, given y(0) = 3. Separate: dy/y = 2x dx. Integrate: ln|y| = x² + C. Exponentiate: y = Ae^(x²). Apply the initial condition: 3 = Ae⁰ = A, so A = 3. Solution: y = 3e^(x²). Check: dy/dx = 3e^(x²)·2x = 6xe^(x²) = 2x(3e^(x²)) = 2xy — matching the original equation.

Worked Example — Modeling Radioactive Decay

A radioactive sample decays according to dy/dt = −0.05y (grams per year), starting at y(0) = 200 grams. By the derived model, y(t) = 200e^(−0.05t). After 10 years: y(10) = 200e^(−0.5) ≈ 200(0.60653) ≈ 121.3 grams. Find the half-life by solving 200e^(−0.05t) = 100: e^(−0.05t) = 0.5, so −0.05t = ln(0.5) = −ln(2) ≈ −0.6931, giving t = 0.6931/0.05 ≈ 13.86 years.

Tip

Apply the initial condition only after fully solving for y — solving y = Ae^(kx) for A using y(0) = y₀ is much simpler than trying to track the initial condition through the integration steps themselves.

Common Mistakes

Forgetting the constant of integration when integrating one side, especially when it's later exponentiated away into a multiplicative constant.
The constant C in ln|y| = kx + C becomes a genuinely different multiplicative constant, A = e^C, after exponentiating — dropping C early loses the very constant an initial condition is needed to determine.
Separating variables incorrectly, leaving an x mixed in with the y-side or vice versa.
Every term has to end up purely in x (with dx) or purely in y (with dy) before integrating — a mixed term signals the equation either needs more algebra first or isn't separable as written.

Key Takeaways

A separable differential equation can be rewritten with all y-terms (and dy) on one side and all x-terms (and dx) on the other, then integrated separately.
dy/dx = ky solves to y = y₀e^(kx), the exponential growth/decay model, derived directly rather than assumed.
An initial condition is applied after solving the general form for y, pinning down the constant of integration as a specific number.

Summary

This closes Unit 7: slope fields and separable equations solve for a function from its own rate of change. Unit 8 returns to the definite integral's other applications — quantities built directly from area, starting with a function's average value.

Track your progress

← Slope Fields Back to Unit 7: Differential Equations