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Separation of Variables

The most frequent FRQ type in AP Calculus. It involves isolating all yy terms with dydy and all xx terms with dxdx using algebraic manipulation before integrating both sides.

Core Theorem
If dydx=g(x)h(y)\frac{dy}{dx} = g(x)h(y), then 1h(y)dy=g(x)dx\frac{1}{h(y)} dy = g(x) dx. Integrate both sides: ∫1h(y)dy=∫g(x)dx\int \frac{1}{h(y)} dy = \int g(x) dx.
Step-by-Step SOP
  1. 1

    Move (Separate)

    Use multiplication/division to get yy terms on the left and xx terms on the right.
  2. 2

    Integrate

    Anti-differentiate both sides and add +C+C to the xx-side.
  3. 3

    Calculate C

    Use the initial condition (x0,y0)(x_0, y_0) to solve for CC before doing complex algebra.
  4. 4

    Isolate y

    Rearrange the equation to the form y=f(x)y = f(x) if required.

Practice Exercises

Example 01Easy
Solve dydx=2xy\frac{dy}{dx} = 2xy given the initial condition y(0)=5y(0) = 5.
NEED A HINT?
Divide by yy and multiply by dxdx. Remember that ∫1ydy=ln⁡∣y∣\int \frac{1}{y} dy = \ln|y|.
SHOW DETAILED EXPLANATION

Step 1: Separate and Integrate

1ydy=2xdx  ⟹  ln⁡∣y∣=x2+C\frac{1}{y} dy = 2x dx \implies \ln|y| = x^2 + C

Step 2: Find C immediately

Plug in (0,5)  ⟹  ln⁡(5)=02+C  ⟹  C=ln⁡(5)(0, 5) \implies \ln(5) = 0^2 + C \implies C = \ln(5).

Step 3: Solve for y

ln⁡∣y∣=x2+ln⁡(5)  ⟹  ∣y∣=ex2+ln⁡(5)=ex2⋅eln⁡(5)  ⟹  y=5ex2\ln|y| = x^2 + \ln(5) \implies |y| = e^{x^2 + \ln(5)} = e^{x^2} \cdot e^{\ln(5)} \implies y = 5e^{x^2}.
Example 02Medium
Solve dydx=xy\frac{dy}{dx} = \frac{x}{y} given the curve passes through (0,2)(0, 2).
NEED A HINT?
Cross-multiply to separate the variables.
SHOW DETAILED EXPLANATION

Step 1: Separate

y dy=x dxy \, dy = x \, dx

Step 2: Integrate

∫y dy=∫x dx  ⟹  12y2=12x2+C\int y \, dy = \int x \, dx \implies \frac{1}{2}y^2 = \frac{1}{2}x^2 + C

Step 3: Solve for C

Substitute (0,2)  ⟹  12(2)2=12(0)2+C  ⟹  2=C(0, 2) \implies \frac{1}{2}(2)^2 = \frac{1}{2}(0)^2 + C \implies 2 = C.

Step 4: Isolate y

12y2=12x2+2  ⟹  y2=x2+4\frac{1}{2}y^2 = \frac{1}{2}x^2 + 2 \implies y^2 = x^2 + 4. Since y(0)=2y(0)=2 is positive, y=x2+4y = \sqrt{x^2 + 4}.
Common Pitfalls
  • ⚠
    The '+C' TrapForgetting +C+C during the integration step usually results in losing 3 to 4 points out of 5 on an AP FRQ. It must be added immediately after integrating.
  • ⚠
    Exponential Constant LogicIn dy/dx=kydy/dx = ky problems, ekt+Ce^{kt+C} becomes CektCe^{kt}. Don't forget that the CC shifts from the exponent to a coefficient.
  • ⚠
    Domain & SignWhen you have y2=dotsy^2 = dots, you must choose between the positive or negative root based on the yy-value of your initial condition.
Gary Chang

Gary Chang

Calculus Educator

5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

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