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Initial Conditions &
Separation of Variables

Learn to solve differential equations by separating variables and using initial conditions to find the one particular solution that fits.

AP Calculus BC

What Is a Differential Equation?

A differential equation (DE) is any equation that contains a function and one or more of its derivatives. It describes a relationship between a quantity and its rate of change.

Solving a DE means finding the function \(y(x)\) that satisfies the equation. The solution is not a number: it is an entire function or family of functions.

General vs. Particular Solutions

General Solution
Contains an arbitrary constant \(C\). Represents a family of curves: \(y = x^3 + C\).
Particular Solution
A single curve from the family, found by using an initial condition to determine \(C\): \(y = x^3 + 2\).
x y (0, 2) C = 0 C = 2 C = −2

What Are Initial Conditions?

An initial condition is a known point \((x_0, y_0)\) that the solution must pass through. It pins down the constant \(C\) so you get exactly one curve instead of infinitely many.

Without the initial condition, \(C\) stays unknown and you only have the general solution.

The Technique: Separation of Variables

Separation of variables is a method for solving DEs of the form \(\dfrac{dy}{dx} = f(x)\,g(y)\). The key idea: rearrange so that all \(y\)-terms are on one side and all \(x\)-terms on the other, then integrate both sides.

Step 1 Separate
\(\frac{1}{g(y)}\,dy = f(x)\,dx\)
Step 2 Integrate
\(\int \frac{1}{g(y)}\,dy = \int f(x)\,dx\)
Step 3 Apply IC
Solve for \(C\)

The Formal Method

This works only when \(\dfrac{dy}{dx}\) can be written as a product of a function of \(x\) alone and a function of \(y\) alone. Not every DE is separable.

Separable vs. Non-Separable

Equation Separable? Why
\(\dfrac{dy}{dx} = x^2 y\) ✓ Yes \(f(x)=x^2,\ g(y)=y\)
\(\dfrac{dy}{dx} = e^{x+y}\) ✓ Yes \(e^{x+y}=e^x \cdot e^y\)
\(\dfrac{dy}{dx} = x + y\) ✕ No Sum, not product
\(\dfrac{dy}{dx} = \sin(xy)\) ✕ No Cannot factor into \(f(x)\cdot g(y)\)

How to Apply the Method

1
Check separability. Confirm \(\dfrac{dy}{dx}\) can be written as \(f(x)\cdot g(y)\).
2
Separate variables. Move all \(y\)-terms (including \(dy\)) to one side, all \(x\)-terms (including \(dx\)) to the other.
3
Integrate both sides. Don't forget the constant \(+C\) on one side.
4
Solve for \(y\) explicitly if possible.
5
Apply the initial condition \(y(x_0)=y_0\) to find the value of \(C\).

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: \(4 = (1)^2 + C \implies C = 3\)

Example 2: Products of Functions

1
Separate: \(\dfrac{1}{y}\,dy = 6x^2\,dx\)
2
Integrate: \(\ln|y| = 2x^3 + C\)
3
Exponentiate: \(y = Ae^{2x^3}\) where \(A = e^C\)
4
Apply IC: \(3 = Ae^{0} = A\), so \(A = 3\)

Visualizing \(y = 3e^{2x^3}\)

x y 0.5 1 −0.5 3 (0, 3)

Example 3: Implicit Solutions

1
Separate: \(y\,dy = x\,dx\)
2
Integrate: \(\dfrac{y^2}{2} = \dfrac{x^2}{2} + C\), or equivalently \(y^2 - x^2 = K\)
3
Apply IC: \(16 - 9 = K \implies K = 7\)
4
Solve for \(y\): Since \(y(3)=4 > 0\), take the positive root.

Why the \(+C\) Matters

Every antiderivative has a \(+C\) because differentiation destroys constant information. The constant of integration represents the vertical shift of the solution curve.

Before IC
Infinite family of curves, each shifted vertically by a different \(C\). All valid solutions to the DE.
After IC
Exactly one curve survives: the one that passes through the given point \((x_0, y_0)\).

Common Mistakes to Avoid

✕ Forgetting \(+C\)
Omitting the constant after integration. Without \(C\), you cannot find the particular solution.
✕ Wrong Separation
Leaving mixed \(x\) and \(y\) terms on the same side. Every \(y\) must go with \(dy\), every \(x\) with \(dx\).
✕ Applying IC Too Early
Plugging in the initial condition before integrating. Always integrate first, then apply the condition to find \(C\).
✕ Sign Errors with \(\ln\)
When exponentiating \(\ln|y|=\ldots+C\), remember \(y=\pm e^C \cdot e^{\ldots}\). Use the IC to determine the sign.
✕ Dividing by Zero
When dividing by \(g(y)\), check that \(g(y)\neq 0\). Solutions where \(g(y)=0\) can be lost (singular solutions).

Summary at a Glance

Concept Key Idea Watch Out
Separable DE \(\dfrac{dy}{dx}=f(x)\cdot g(y)\) Must be a product, not a sum
General Solution Contains arbitrary \(C\) Don't forget \(+C\)
Initial Condition \(y(x_0)=y_0\) pins down \(C\) Apply after integrating
Particular Solution One specific function Check domain and sign
Implicit vs. Explicit Sometimes \(y\) stays inside an equation Use IC to pick correct branch

Question 1 of 5

Solve \(\dfrac{dy}{dx} = 4x\) with \(y(0) = 7\). What is the particular solution?

Question 2 of 5

Which of these differential equations is separable?

Question 3 of 5

Solve \(\dfrac{dy}{dx} = y\) with \(y(0) = 5\). What is the particular solution?

Question 4 of 5

Given \(\dfrac{dy}{dx} = \dfrac{2x}{y}\) and \(y(0) = 3\), find \(y\) when \(x = 2\).

Question 5 of 5

A student solves \(\dfrac{dy}{dx} = 3x^2 y\) and writes the general solution as \(y = e^{x^3}\). What is their error?