Unit 7.6 – Finding General Solutions Using Separation of Variables

AP® Calculus AB & BC | Solving Differential Equations Algebraically

Why This Matters: Separation of variables is THE fundamental technique for solving differential equations algebraically! Up to now, you've verified solutions and visualized them with slope fields. Now you'll actually SOLVE them! This method works when you can separate all \(x\)'s on one side and all \(y\)'s on the other. It's essential for both AB and BC, appears on virtually every AP® exam, and gives exact solutions (not approximations). Master this and you've unlocked the power to solve real-world problems!

🎯 What is Separation of Variables?

DEFINITION

Separation of variables is a method for solving differential equations by:

Separating all terms with \(y\) (including \(dy\)) to one side
Separating all terms with \(x\) (including \(dx\)) to the other side
Integrating both sides

The General Idea:

\[ \frac{dy}{dx} = f(x) \cdot g(y) \quad \Rightarrow \quad \frac{1}{g(y)} \, dy = f(x) \, dx \]

Then integrate both sides!

✅ When Can We Use Separation of Variables?

A DE is separable if it can be written as:

\[ \frac{dy}{dx} = f(x) \cdot g(y) \]

Key: The right side must be a product of a function of \(x\) alone and a function of \(y\) alone!

Examples of Separable DEs:

\(\frac{dy}{dx} = xy\) → separable (can write as \(x \cdot y\))
\(\frac{dy}{dx} = ky\) → separable (can write as \(k \cdot y\))
\(\frac{dy}{dx} = \frac{x}{y}\) → separable (can write as \(x \cdot \frac{1}{y}\))
\(\frac{dy}{dx} = y^2 - 1\) → separable (write as \(1 \cdot (y^2-1)\))

NOT Separable:

\(\frac{dy}{dx} = x + y\) → NOT separable (can't factor into \(f(x) \cdot g(y)\))
\(\frac{dy}{dx} = xy + x\) → NOT separable

📋 The Separation of Variables Method

Step-by-Step Process

The 6-Step Method:

Check if separable: Can you write as \(\frac{dy}{dx} = f(x) \cdot g(y)\)?
Separate variables: Get all \(y\)'s and \(dy\) on one side, all \(x\)'s and \(dx\) on other
\[ \frac{1}{g(y)} \, dy = f(x) \, dx \]
Integrate both sides:
\[ \int \frac{1}{g(y)} \, dy = \int f(x) \, dx \]
Don't forget +C: Add constant on ONE side only
Solve for \(y\): If possible (general solution)
Apply initial condition: If given, to find particular solution

📖 Comprehensive Worked Examples

Example 1: Basic Exponential Growth

Problem: Solve \(\frac{dy}{dx} = ky\) (where \(k\) is a constant)

Solution:

Step 1: Check if separable

Yes! Can write as \(k \cdot y\) (function of \(x\) times function of \(y\))

Step 2: Separate variables

\[ \frac{1}{y} \, dy = k \, dx \]

Step 3: Integrate both sides

\[ \int \frac{1}{y} \, dy = \int k \, dx \]

\[ \ln|y| = kx + C \]

Step 5: Solve for \(y\)

\[ |y| = e^{kx + C} = e^C \cdot e^{kx} \]

Let \(A = \pm e^C\) (a constant):

\[ y = Ae^{kx} \]

GENERAL SOLUTION: \(y = Ae^{kx}\)

Example 2: With Initial Condition

Problem: Solve \(\frac{dy}{dx} = 2xy\) with \(y(0) = 3\)

Solution:

Steps 1-2: Separate

\[ \frac{1}{y} \, dy = 2x \, dx \]

Step 3: Integrate

\[ \int \frac{1}{y} \, dy = \int 2x \, dx \]

\[ \ln|y| = x^2 + C \]

Step 5: Solve for \(y\)

\[ y = Ae^{x^2} \]

Step 6: Apply initial condition \(y(0) = 3\)

\[ 3 = Ae^{0} = A \]

So \(A = 3\)

PARTICULAR SOLUTION: \(y = 3e^{x^2}\)

Example 3: Fraction Form

Problem: Solve \(\frac{dy}{dx} = \frac{x}{y}\)

Solution:

Steps 1-2: Separate

\[ y \, dy = x \, dx \]

Step 3: Integrate

\[ \int y \, dy = \int x \, dx \]

\[ \frac{y^2}{2} = \frac{x^2}{2} + C \]

Step 5: Solve for \(y\) (optional)

Multiply by 2:

\[ y^2 = x^2 + 2C \]

Let \(K = 2C\):

\[ y^2 = x^2 + K \]

Or: \(y = \pm\sqrt{x^2 + K}\)

GENERAL SOLUTION: \(y^2 = x^2 + K\) or \(y = \pm\sqrt{x^2 + K}\)

Example 4: More Complex

Problem: Solve \(\frac{dy}{dx} = \frac{y^2 - 1}{x^2}\)

Solution:

Steps 1-2: Separate

\[ \frac{1}{y^2 - 1} \, dy = \frac{1}{x^2} \, dx \]

Step 3: Integrate (use partial fractions for left side)

For \(\frac{1}{y^2-1} = \frac{1}{(y-1)(y+1)}\), use partial fractions:

\[ \frac{1}{y^2-1} = \frac{1/2}{y-1} - \frac{1/2}{y+1} \]

\[ \int \left(\frac{1/2}{y-1} - \frac{1/2}{y+1}\right) dy = \int x^{-2} \, dx \]

\[ \frac{1}{2}\ln|y-1| - \frac{1}{2}\ln|y+1| = -\frac{1}{x} + C \]

Simplify:

\[ \frac{1}{2}\ln\left|\frac{y-1}{y+1}\right| = -\frac{1}{x} + C \]

GENERAL SOLUTION (implicit form): \(\frac{1}{2}\ln\left|\frac{y-1}{y+1}\right| = -\frac{1}{x} + C\)

🔍 Common Separable DE Types

Frequently Encountered Forms

Common Separable Differential Equations
Type	Form	General Solution
Exponential Growth/Decay	\(\frac{dy}{dx} = ky\)	\(y = Ae^{kx}\)
Product Form	\(\frac{dy}{dx} = xy\)	\(y = Ae^{x^2/2}\)
Quotient Form	\(\frac{dy}{dx} = \frac{x}{y}\)	\(y^2 = x^2 + C\)
Pure \(x\) Function	\(\frac{dy}{dx} = f(x)\)	\(y = \int f(x)\,dx + C\)
Pure \(y\) Function	\(\frac{dy}{dx} = g(y)\)	\(\int \frac{1}{g(y)}\,dy = x + C\)

🎲 General Solution vs Particular Solution

Understanding the Difference:

General Solution:

Contains arbitrary constant(s) (usually \(C\) or \(A\))
Represents family of all solutions
Example: \(y = Ae^{2x}\)

Particular Solution:

Specific solution (no arbitrary constants)
Found by applying initial condition
Example: \(y = 5e^{2x}\) (when \(y(0) = 5\))

💡 Essential Tips & Tricks

✅ Success Strategies:

Check if separable first: Don't waste time if it can't be separated
Get all \(dy\) with \(y\): And all \(dx\) with \(x\)
Add +C on ONE side only: Not both sides
Simplify before solving for \(y\): Makes algebra easier
Don't forget absolute values: In \(\ln|y|\)
Constants can be renamed: \(e^C\) becomes \(A\), etc.
Check your answer: Differentiate and verify it satisfies the DE

🔥 Integration Tricks:

\(\int \frac{1}{y}\,dy = \ln|y| + C\): Most common integral
For \(\frac{1}{y^2-a^2}\): Use partial fractions
For \(\frac{1}{y^2+a^2}\): Use \(\arctan\)
Factor before integrating: Might simplify
U-substitution: May be needed on one or both sides

💡 Constant Manipulation Trick:

When you get \(\ln|y| = f(x) + C\), to solve for \(y\):

\(|y| = e^{f(x)+C} = e^C \cdot e^{f(x)}\)
Let \(A = \pm e^C\) (this is just another constant)
\(y = Ae^{f(x)}\)

This is cleaner than keeping \(\pm e^C\) around!

❌ Common Mistakes to Avoid

Mistake 1: Adding +C to both sides (only add to ONE side)
Mistake 2: Forgetting absolute value: \(\int \frac{1}{y}\,dy = \ln|y|\), not \(\ln y\)
Mistake 3: Not checking if equation is separable before starting
Mistake 4: Algebra errors when separating variables
Mistake 5: Integration errors (especially with fractions)
Mistake 6: Forgetting to apply initial condition
Mistake 7: Sign errors when moving terms
Mistake 8: Not simplifying before solving for \(y\)
Mistake 9: Losing negative signs or constants
Mistake 10: Not checking answer (always verify!)

📝 Practice Problems

Solve these differential equations:

\(\frac{dy}{dx} = 3y\), \(y(0) = 2\)
\(\frac{dy}{dx} = xy^2\)
\(\frac{dy}{dx} = \frac{y}{x}\), \(y(1) = 4\)
\(\frac{dy}{dx} = e^{x-y}\)

Answers:

\(y = 2e^{3x}\)
\(y = \frac{-1}{x^2/2 + C}\) or \(y = \frac{-2}{x^2 + K}\)
\(y = 4x\)
\(e^y = e^x + C\) or \(y = \ln(e^x + C)\)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Clear separation: Show work getting variables separated
Integration step: Show integral signs and integration work
Include +C: On one side after integrating
Solve for \(y\): Unless question asks for implicit form
Apply initial condition: Show substitution to find constant
Final answer clearly stated: Box or underline it
Verification (if time): Shows understanding

💯 Exam Strategy:

Read problem completely—what are you finding?
Check if separable—if not, skip or use other method
Separate variables carefully—show your work
Integrate both sides—show integral signs
Solve for \(y\) if possible
If initial condition given, apply it to find constant
State final answer clearly
If time permits, verify by differentiating

⚡ Quick Reference Guide

SEPARATION OF VARIABLES ESSENTIALS

The Method:

Check: Is \(\frac{dy}{dx} = f(x) \cdot g(y)\)?
Separate: \(\frac{1}{g(y)}\,dy = f(x)\,dx\)
Integrate: \(\int \frac{1}{g(y)}\,dy = \int f(x)\,dx\)
Add +C (one side only!)
Solve for \(y\)
Apply initial condition if given

Key Formulas:

\(\frac{dy}{dx} = ky\): Solution is \(y = Ae^{kx}\)
\(\int \frac{1}{y}\,dy = \ln|y| + C\)
General to Particular: Use initial condition to find constant

Master Separation of Variables! This method solves differential equations algebraically when the DE is separable: \(\frac{dy}{dx} = f(x) \cdot g(y)\). The process: (1) Check if separable, (2) Separate variables: get all \(y\)'s and \(dy\) on one side, all \(x\)'s and \(dx\) on other: \(\frac{1}{g(y)}\,dy = f(x)\,dx\), (3) Integrate both sides, (4) Add +C to ONE side only, (5) Solve for \(y\) to get general solution, (6) Apply initial condition to find particular solution. Most common: \(\frac{dy}{dx} = ky\) gives \(y = Ae^{kx}\). Remember \(\int \frac{1}{y}\,dy = \ln|y| + C\). When solving for \(y\), rename constants: \(e^C\) becomes \(A\). Always verify by differentiating your answer. Common mistakes: adding +C to both sides, forgetting absolute values, algebra errors during separation. Show ALL work on AP® exams—separation step, integration with integral signs, +C, solving for \(y\), applying initial condition. This appears on virtually every AP® exam! Practice until automatic! 🎯✨