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Separation of Variables

The most frequent FRQ type in AP Calculus. It involves isolating all $y$ terms with $dy$ and all $x$ terms with $dx$ 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.
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