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Initial Conditions &
Separation of Variables

Learn to solve differential equations by separating variables and applying initial conditions to find particular solutions.

AP Calculus AB

What Is a Differential Equation?

A differential equation is an equation that relates a function to its own derivatives. It tells us how a quantity changes rather than what the quantity is.

Example
\(\dfrac{dy}{dx} = 3x^2\) says the rate of change of \(y\) equals \(3x^2\).
Goal
Find the function \(y = f(x)\) that satisfies the equation.

General vs. Particular Solutions

When you integrate a differential equation, you get a family of curves differing by a constant \(C\). An initial condition pins down exactly one curve.

x y
\(y(0)=0\)
General family Particular solution

Why Do Initial Conditions Matter?

Without an initial condition, we only have a general solution containing an arbitrary constant \(C\). The initial condition gives us a specific point \((x_0, y_0)\) on the curve, which lets us solve for \(C\).

General Solution
\(y = x^3 + C\): infinitely many curves
Initial Condition
\(y(1) = 5\): a specific data point
Particular Solution
\(y = x^3 + 4\): exactly one curve

The Separation of Variables Technique

Separation of variables works when a differential equation can be written so that all \(y\)-terms are on one side and all \(x\)-terms are on the other:

Then integrate both sides independently:

When Can We Separate?

Separation of variables requires the right-hand side to be a product (or quotient) of a function of \(x\) alone and a function of \(y\) alone.

✓ Separable
\(\dfrac{dy}{dx} = x^2 y\)

\(\dfrac{dy}{dx} = \dfrac{e^x}{y^2}\)

\(\dfrac{dy}{dx} = (1+x)(1+y)\)
✕ Not Separable
\(\dfrac{dy}{dx} = x + y\)

\(\dfrac{dy}{dx} = xy + x^2\)

\(\dfrac{dy}{dx} = \sin(x + y)\)

General vs. Particular: Side by Side

General Solution Particular Solution
Contains \(C\)? Yes: arbitrary constant No: \(C\) is determined
Number of curves Infinitely many Exactly one
Uses initial condition? Not yet Yes: \(y(x_0) = y_0\)
Example \(y = x^2 + C\) \(y = x^2 + 3\)

How to Solve: Step by Step

1
Separate: rewrite \(\dfrac{dy}{dx} = f(x) \cdot g(y)\) as \(\dfrac{dy}{g(y)} = f(x)\,dx\).
2
Integrate both sides: \(\displaystyle\int \frac{dy}{g(y)} = \int f(x)\,dx + C\).
3
Solve for \(y\) explicitly if possible (isolate \(y\)).
4
Apply initial condition: substitute \((x_0, y_0)\) to find \(C\).
5
Write the particular solution with the determined \(C\).

Example 1: Basic Separation

1
Separate: \(dy = 2x\,dx\)
2
Integrate: \(\displaystyle\int dy = \int 2x\,dx \implies y = x^2 + C\)
3
Apply IC: \(y(0) = 3 \implies 3 = 0 + C \implies C = 3\)

Example 2: Involving \(y\) on Both Sides

1
Separate: \(y\,dy = x\,dx\)
2
Integrate: \(\dfrac{y^2}{2} = \dfrac{x^2}{2} + C\)
3
Apply IC: \(\dfrac{4}{2} = 0 + C \implies C = 2\)
4
Solve for \(y\): \(y^2 = x^2 + 4 \implies y = \sqrt{x^2 + 4}\) (positive since \(y(0)=2>0\))

Example 3: Exponential Growth

1
Separate: \(\dfrac{dy}{y} = 3\,dx\)
2
Integrate: \(\ln|y| = 3x + C_1 \implies y = Ae^{3x}\) where \(A = \pm e^{C_1}\)
3
Apply IC: \(y(0) = 5 \implies A \cdot e^0 = 5 \implies A = 5\)

Visualizing \(y = 5e^{3x}\)

x y 0 0.5 1.0 1.5 0 20 40 60 80 100 (0, 5)
\(y = 5e^{3x}\)
\(A = 8\)
\(A = 2\)

Example 4: More Involved Separation

1
Separate: \((y+1)\,dy = 2x\,dx\)
2
Integrate: \(\dfrac{y^2}{2} + y = x^2 + C\)
3
Apply IC: \(\dfrac{1}{2} + 1 = 0 + C \implies C = \dfrac{3}{2}\)
4
Particular solution: \(\dfrac{y^2}{2} + y = x^2 + \dfrac{3}{2}\)

Common Mistakes to Avoid

✕ Forgetting \(+C\)
Every indefinite integral produces a constant. Without \(C\), you cannot apply the initial condition.
✕ Wrong separation
Treating \(\dfrac{dy}{dx}\) as a fraction is valid here, but forgetting to move all \(y\)-terms to one side leads to errors.
✕ Dropping absolute value
When integrating \(\dfrac{1}{y}\,dy\), the result is \(\ln|y|\), not \(\ln y\). The initial condition determines the sign.

Summary: The Complete Process

Step Action Result
1. Check Is \(\dfrac{dy}{dx} = f(x) \cdot g(y)\)? Confirm separability
2. Separate Move \(y\)-terms left, \(x\)-terms right \(\dfrac{dy}{g(y)} = f(x)\,dx\)
3. Integrate Antidifferentiate both sides General solution with \(C\)
4. Apply IC Substitute \((x_0, y_0)\) Determine \(C\)
5. Write Plug \(C\) back in Particular solution

Question 1 of 5

Solve \(\dfrac{dy}{dx} = 4x\) with \(y(1) = 7\). What is the particular solution?

Question 2 of 5

Which equation is separable?

Question 3 of 5

Solve \(\dfrac{dy}{dx} = 2y\) with \(y(0) = 3\). What is the particular solution?

Question 4 of 5

When solving \(\dfrac{dy}{dx} = \dfrac{x}{y}\) with \(y(3) = 4\), after integrating you get \(\dfrac{y^2}{2} = \dfrac{x^2}{2} + C\). What is \(C\)?

Question 5 of 5

What is the first step in solving \(\dfrac{dy}{dx} = e^x \cdot y^2\) by separation of variables?