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Separation of
Variables

Master the technique of separating and integrating to solve first-order differential equations and find their general solutions.

AP Calculus BC

What Is a Differential Equation?

A differential equation (DE) is an equation that relates a function to one or more of its derivatives. In AP Calculus BC, we focus on first-order ordinary DEs: equations involving \(\frac{dy}{dx}\) and the variables \(x\) and \(y\).

A general solution contains an arbitrary constant \(C\) and represents a family of curves. A particular solution uses an initial condition to pin down a single curve.

Why Separation of Variables?

Not every DE can be solved in closed form. Separation of variables is one of the most powerful and accessible methods. It works whenever we can rewrite the equation so that each side involves only one variable.

When It Works
\(\frac{dy}{dx}\) can be expressed as a product \(f(x) \cdot g(y)\).
When It Fails
Terms mix \(x\) and \(y\) inseparably, e.g. \(\frac{dy}{dx} = x + y\).

A Family of Solution Curves

Solving \(\frac{dy}{dx} = 2x\) gives \(y = x^2 + C\). Each value of \(C\) produces a different parabola: the general solution is the entire family.

x y -2 2 C = 2 C = 0 C = -2

The Separation Technique

Given a separable DE of the form:

The constant of integration \(C\) appears once (combining both integration constants). The result is the general solution: an implicit or explicit relationship between \(x\) and \(y\).

Is the DE Separable?

Given \(\frac{dy}{dx} = h(x,y)\)
Can you write \(h(x,y) = f(x) \cdot g(y)\)?
 Yes No Separable
1. Move \(g(y)\) to left
2. Move \(f(x)\) to right
3. Integrate both sides
 Not Separable Try another method: integrating factor, exact equations, etc.

Separable vs. Linear DEs

Separable First-Order Linear
Standard Form \(\frac{dy}{dx} = f(x)\,g(y)\) \(\frac{dy}{dx} + P(x)\,y = Q(x)\)
Method Separate and integrate Integrating factor \(\mu(x)\)
Overlap? Some DEs are both separable and linear
Example \(\frac{dy}{dx} = xy^2\) \(\frac{dy}{dx} + 2y = e^x\)
AP BC Focus ✓ Primary method Awareness only

How to Apply the Method

1
Identify: Confirm the DE is separable. Write \(\frac{dy}{dx} = f(x) \cdot g(y)\).
2
Separate: Move all \(y\)-terms (including \(dy\)) to the left, all \(x\)-terms (including \(dx\)) to the right.
3
Integrate: Compute \(\int \frac{1}{g(y)}\,dy = \int f(x)\,dx\). Add \(+ C\) on one side only.
4
Solve (if possible): Isolate \(y\) explicitly. If not possible, leave the answer in implicit form.
5
Verify: Differentiate your solution and confirm it satisfies the original DE.

Example 1: Basic Separation

1
Separate: \(\dfrac{1}{y}\,dy = 3x^2\,dx\)
2
Integrate: \(\displaystyle\int \frac{1}{y}\,dy = \int 3x^2\,dx\) gives \(\ln|y| = x^3 + C_1\)
3
Solve for \(y\): \(|y| = e^{x^3 + C_1} = e^{C_1} \cdot e^{x^3}\). Let \(A = \pm e^{C_1}\).

Example 2: Rational Separation

1
Separate: \(y^2\,dy = x\,dx\)
2
Integrate: \(\displaystyle\int y^2\,dy = \int x\,dx\) gives \(\dfrac{y^3}{3} = \dfrac{x^2}{2} + C\)
3
Solve for \(y\): \(y^3 = \dfrac{3x^2}{2} + K\), so \(y = \left(\dfrac{3x^2}{2} + K\right)^{1/3}\)

Example 3: Trig and Exponential

1
Separate: \(\dfrac{1}{\cos y}\,dy = e^x\,dx\), i.e. \(\sec y\,dy = e^x\,dx\)
2
Integrate: \(\displaystyle\int \sec y\,dy = \int e^x\,dx\) gives \(\ln|\sec y + \tan y| = e^x + C\)
3
Result: The general solution is implicit: \(\ln|\sec y + \tan y| = e^x + C\). Not every DE yields an explicit \(y = \ldots\) form.

The Role of the Constant \(C\)

When integrating both sides, each integral produces its own constant. We combine them into a single constant \(C\) on one side:

General Solution
Contains \(C\): represents an infinite family of curves.
Particular Solution
An initial condition \(y(x_0) = y_0\) pins down a unique \(C\).

Handling Absolute Values

When \(\int \frac{1}{y}\,dy = \ln|y|\) appears, the absolute value leads to a \(\pm\) when exponentiating. The standard trick:

1
\(\ln|y| = F(x) + C_1\) implies \(|y| = e^{F(x) + C_1} = e^{C_1} \cdot e^{F(x)}\).
2
Since \(|y| = \pm y\), we write \(y = \pm e^{C_1} \cdot e^{F(x)}\).
3
Replace \(\pm e^{C_1}\) with a single new constant \(A \neq 0\). Result: \(y = A e^{F(x)}\).

Implicit vs. Explicit Solutions

Explicit
\(y = f(x)\). Directly gives \(y\) as a function of \(x\). Preferred when algebra is clean.
Implicit
\(G(y) = F(x) + C\). Acceptable when isolating \(y\) is impossible or creates messy expressions.

Examples of naturally implicit results:

Common Mistakes to Avoid

✕ Forgetting \(+ C\)
A general solution must include the arbitrary constant. Omitting it gives only one curve, not the family.
✕ Two Constants
Writing \(C_1\) on the left and \(C_2\) on the right, then never combining. Use one \(C\) on one side.
✕ Dropping \(|y|\)
Writing \(\ln y\) instead of \(\ln|y|\) can miss negative solutions. The \(\pm\) matters for the general case.
✕ Bad Algebra in Separation
Dividing by \(g(y)\) when \(g(y) = 0\) is possible. Always check if the zero case gives an extra solution.
✕ Incorrect Integration
Rushing the antiderivative. Double-check: \(\int \sec y\,dy = \ln|\sec y + \tan y|\), not \(\ln|\sec y|\).

Method Summary

Step Action Result
1. Identify Write as \(f(x) \cdot g(y)\) Confirm separable
2. Separate \(\frac{dy}{g(y)} = f(x)\,dx\) Variables isolated
3. Integrate \(\int\) both sides \(G(y) = F(x) + C\)
4. Solve Isolate \(y\) if possible Explicit or implicit
5. Verify Differentiate and plug back Matches original DE

Question 1 of 5

Which of the following DEs is separable?

Question 2 of 5

Solve \(\dfrac{dy}{dx} = 2xy\). What is the general solution?

Question 3 of 5

After separating and integrating \(\dfrac{dy}{dx} = \dfrac{y}{x}\), which result is correct?

Question 4 of 5

When solving \(\dfrac{dy}{dx} = y^2 \sin x\) by separation, what should you integrate on the left side?

Question 5 of 5

A student solves a separable DE and writes \(\dfrac{y^3}{3} = 2x + C_1\) and \(\dfrac{y^3}{3} + C_2 = 2x\). What error did they make?