Separation of Variables
- Separate a differential equation so each variable is on its own side
- Integrate both sides and solve for the general solution
- Apply an initial condition to find the particular solution
Getting the variables apart
A differential equation is separable if it can be written so all the y’s (with dy) are on one side and all the x’s (with dx) on the other: dy/dx = g(x)·h(y) rearranges to (1/h(y)) dy = g(x) dx. Once separated, you can integrate each side with respect to its own variable. Not every equation separates, but many AB problems are designed to.
Integrate both sides, then solve
After separating, integrate both sides. A single constant of integration C (combine the two) appears. Then solve for y explicitly if possible — often you must exponentiate to undo a logarithm. The result with the free C is the general solution, a whole family of curves; each value of C gives a different member.
Use the initial condition for C
A particular solution is pinned down by an initial condition like y(0) = 5. Substitute the given point into the general solution and solve for C. It is usually cleaner to find C right after integrating (before solving for y) to avoid dragging the constant through the algebra. Report the final solution with the specific C in place.
Solve dy/dx = 2xy with initial condition y(0) = 3.
- 1.Separate: (1/y) dy = 2x dx.
- 2.Integrate both sides: ln|y| = x² + C.
- 3.Solve for y by exponentiating: |y| = e^(x² + C) = e^C·e^(x²), so y = A·e^(x²) where A = ±e^C.
- 4.Apply y(0) = 3: 3 = A·e⁰ = A, so A = 3.
Do not forget the constant of integration, and add it right after integrating — before you solve for y. Introducing C too late, or leaving it out entirely, is the most common way separable-equation problems go wrong.
To solve dy/dx = x/y by separation of variables, the correct separated form is:
After separating and integrating dy/dx = 3x², you obtain y = x³ + C. If y(1) = 5, what is C?
Free-response separable-equation problems award points for the separated equation, the antiderivatives of both sides, and finding C from the initial condition. Show each stage explicitly — a correct final formula with no work in between leaves points on the table.
Answer the 2 checkpoints as you read.
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