Separation of Variables

You’ll be able to

Solve separable differential equations by separating and integrating
Apply an initial condition to find the particular solution
Recognize exponential growth and decay as separable models

When the variables can be split apart

A differential equation is separable if it can be rearranged so all the y’s (with dy) sit on one side and all the x’s (with dx) sit on the other: g(y) dy = h(x) dx. Once separated, you integrate both sides. This yields the general solution as a relationship between x and y, complete with a single constant of integration. Separation is the primary analytic technique for the differential equations on the BC exam.

General to particular via the initial condition

Integrating a separable equation leaves a constant C; the solution so far is a whole family of curves. An initial condition — a known point (x₀, y₀) the solution passes through — pins down C. Substitute the point, solve for C, and you have the particular solution. Best practice is to solve for C right after integrating (before simplifying), which avoids sign and algebra errors when you later isolate y.

Exponential growth/decay model

dy/dt = k·y ⟹ y = y₀·e^{kt}

The classic separable equation: rate proportional to amount. k > 0 is growth, k < 0 is decay; y₀ is the initial amount.

Worked example

Solve dy/dx = xy with the initial condition y(0) = 3.

1.Separate the variables: (1/y) dy = x dx.
2.Integrate both sides: ln|y| = x²/2 + C.
3.Exponentiate to solve for y: y = e^{x²/2 + C} = A·e^{x²/2}, where A = e^C.
4.Apply y(0) = 3: 3 = A·e^0 = A, so A = 3 and y = 3·e^{x²/2}.

Answer: The particular solution is y = 3·e^{x²/2}. The initial condition fixed the constant A at 3.

Watch out

Do not forget the constant of integration, and add it before exponentiating. Writing ln|y| = x²/2 and only later inserting C leads to the wrong form; the C must ride through the exponentiation to become the multiplicative constant A.

Checkpoint

Which is the correct first step in solving the separable equation dy/dx = 2xy²?

Checkpoint

A quantity grows so that dP/dt = 0.05P, with P(0) = 200. What is P(t)?

On the exam

On free response you must show the separation and the integration of both sides, then apply the initial condition explicitly. A bare final formula, even if correct, typically earns only partial credit — the graders reward the separated equation and the antiderivatives.

Answer the 2 checkpoints as you read.