7.6 - Finding General Solutions Using Separation of Variables

Review: Implicit Differentiation

Before learning separation of variables, let’s recall implicit differentiation.

When an equation isn’t solved for

y

, meaning it isn’t in the form

y = \ldots

, we differentiate implicitly with respect to

x

Example: Differentiate

x^2 + y^2 = 25

\frac{d}{dx}[x^2 + y^2] = \frac{d}{dx}[25]

2x + 2y\frac{dy}{dx} = 0

Solving for

\frac{dy}{dx}

2y\frac{dy}{dx} = -2x \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{x}{y}

Separation of variables works in reverse. We start with

\frac{dy}{dx}

and integrate to find

y

General vs. Particular Solutions

General solution: contains an arbitrary constant

C

, representing a whole family of solutions.

Particular solution: a specific solution, found by using an initial condition to determine

C

This lesson focuses on general solutions, the ones with the

+C

. Particular solutions come in the next lesson.

What is Separation of Variables?

It’s a technique for solving differential equations by:

Separating all the terms with $y$ (and $dy$ ) onto one side

Separating all the terms with $x$ (and $dx$ ) onto the other side

Integrating both sides

This only works when

\frac{dy}{dx}

can be written as (a function of $x$ ) times (a function of $y$ ). Some differential equations can’t be separated at all. For example,

\frac{dy}{dx} = x + y

is a sum that won’t factor into an

x

part times a

y

part, so separation of variables can’t solve it. For equations like that, use a slope field (or, in BC, Euler’s method) instead.

Leibniz Notation

The notation

\frac{dy}{dx}

is called Leibniz notation, named after Gottfried Leibniz.

While

\frac{dy}{dx}

is not technically a fraction, we can treat it like one for separation of variables.

So we can “multiply both sides by

dx

” to separate the differentials:

\frac{dy}{dx} = f(x) \quad \Rightarrow \quad dy = f(x) \, dx

The Separation of Variables Process

Step 1: Separate the variables. Rearrange so all the

y

terms (with

dy

) are on one side and all the

x

terms (with

dx

) are on the other.

Step 2: Integrate both sides.

\int g(y) \, dy = \int f(x) \, dx

Step 3: Solve for $y$ if you can. Manipulate algebraically to get

y

by itself, giving an explicit solution.

Example 1: Basic Separation

Solve the differential equation

\frac{dy}{dx} = \frac{x^2}{y}

Step 1: Separate the variables

Multiply both sides by

y

y \frac{dy}{dx} = x^2

Treat

\frac{dy}{dx}

as a fraction and multiply both sides by

dx

y \, dy = x^2 \, dx

Step 2: Integrate both sides

\int y \, dy = \int x^2 \, dx

\frac{y^2}{2} + C_1 = \frac{x^3}{3} + C_2

Note: we have two constants, but we can combine them:

\frac{y^2}{2} = \frac{x^3}{3} + C

where

C = C_2 - C_1

is just one combined constant.

Step 3: Solve for $y$ to get the explicit solution

y^2 = \frac{2x^3}{3} + 2C

Since

2C

is still just a constant, we can keep calling it

C

y^2 = \frac{2x^3}{3} + C

y = \pm\sqrt{\frac{2x^3}{3} + C}

Important: don’t forget the

\pm

when taking even roots.

Answer:

y = \pm\sqrt{\frac{2x^3}{3} + C}

Implicit vs. Explicit Solutions

Implicit Solution

An equation relating

x

and

y

that is not solved for $y$ .

Example:

3y^2 - 2x^3 = C

Explicit Solution

y

is isolated on one side.

Example:

y = \pm\sqrt{\frac{2x^3}{3} + C}

Most problems ask for an explicit solution when one is possible.

Example 2: Trigonometric Differential Equation

Solve

y^2 \frac{dy}{dx} = \sin(x)

Step 1: Separate the variables

Multiply both sides by

dx

y^2 \, dy = \sin(x) \, dx

Step 2: Integrate

\int y^2 \, dy = \int \sin(x) \, dx

\frac{y^3}{3} = -\cos(x) + C

From here on, we only write one constant

C

, on the right side, since we’d combine them anyway.

Step 3: Solve for $y$

y^3 = -3\cos(x) + 3C

Again,

3C

is just another constant:

y^3 = -3\cos(x) + C

y = \sqrt[3]{-3\cos(x) + C}

Note: no

\pm

is needed here, because we’re taking a cube root, which is an odd root.

Answer:

y = \sqrt[3]{-3\cos(x) + C}

Example 3: Same Idea, Different Wording

Solve

\frac{dy}{dx} = \frac{\sin(x)}{y^2}

Step 1: Separate

y^2 \, dy = \sin(x) \, dx

This is the exact same separated equation as Example 2, just written differently in the original problem.

Step 2: Integrate

\int y^2 \, dy = \int \sin(x) \, dx

\frac{y^3}{3} = -\cos(x) + C

Step 3: Solve for $y$

y^3 = -3\cos(x) + C

y = \sqrt[3]{-3\cos(x) + C}

Answer:

y = \sqrt[3]{-3\cos(x) + C}

Example 4: Negative Exponent Form

Solve

\frac{dy}{dx} = y^{-2} \cos(x)

Step 1: Separate

Rewrite

y^{-2} = \frac{1}{y^2}

, so:

y^2 \, dy = \cos(x) \, dx

Step 2: Integrate

\int y^2 \, dy = \int \cos(x) \, dx

\frac{y^3}{3} = \sin(x) + C

Step 3: Multiply by 3

y^3 = 3\sin(x) + C

Step 4: Solve for $y$

y = \sqrt[3]{3\sin(x) + C}

Answer:

y = \sqrt[3]{3\sin(x) + C}

Example 5: The Reciprocal Technique

Solve

\frac{dy}{dx} = y^2 \sin(x)

Step 1: Separate

\frac{1}{y^2} \, dy = \sin(x) \, dx

Step 2: Integrate

Rewrite

\frac{1}{y^2} = y^{-2}

, and recall that

\int y^{-2} \, dy = \frac{y^{-1}}{-1} = -\frac{1}{y}

\int y^{-2} \, dy = \int \sin(x) \, dx

-\frac{1}{y} = -\cos(x) + C

Step 3: Multiply by $-1$

\frac{1}{y} = \cos(x) - C

Since a constant can be positive or negative, we just write it as

+C

\frac{1}{y} = \cos(x) + C

Step 4: Take the reciprocal of both sides

y = \frac{1}{\cos(x) + C}

Answer:

y = \frac{1}{\cos(x) + C}

Key point: the constant has to stay in the denominator when you take the reciprocal.

Example 6: Natural Logarithm and Exponentiating

Solve

\frac{dy}{dx} = xy

Step 1: Separate

\frac{1}{y} \, dy = x \, dx

Step 2: Integrate

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

\ln|y| = \frac{x^2}{2} + C

Step 3: Exponentiate both sides

To solve for

y

, raise

e

to both sides:

e^{\ln|y|} = e^{x^2/2 + C}

|y| = e^{x^2/2 + C}

Step 4: Simplify using exponent properties

Recall that

e^{a+b} = e^a \cdot e^b

|y| = e^{x^2/2} \cdot e^C

Dropping the absolute value gives

y = \pm e^C \cdot e^{x^2/2}

. Since

\pm e^C

is just a constant, we call it

C

y = Ce^{x^2/2}

Answer:

y = Ce^{x^2/2}

Important: after exponentiating, the constant moves from the exponent to a multiplicative coefficient out front.

Why We Rewrite $e^{x^2/2 + C}$ as $Ce^{x^2/2}$

Using exponent rules:

e^{x^2/2 + C} = e^{x^2/2} \cdot e^C

Since

C

is an arbitrary constant,

e^C

is an arbitrary positive constant. The

\pm

from dropping the absolute value lets it be negative too, so we just call the whole thing

C

y = Ce^{x^2/2}

This is the form that shows up in multiple choice answers, and it’s the preferred format.

Example 7: Another Exponential Solution

Solve

\frac{dy}{dx} = 3xy

Step 1: Separate

\frac{dy}{y} = 3x \, dx

Step 2: Integrate

\ln|y| = \frac{3x^2}{2} + C

Step 3: Exponentiate

|y| = e^{3x^2/2 + C} = e^{3x^2/2} \cdot e^C

y = Ce^{3x^2/2}

Answer:

y = Ce^{3x^2/2}

Example 8: Factoring Before Separating

Solve

\frac{dy}{dx} + 2xy = 4x

The Wrong Way

If you just subtract

2xy

\frac{dy}{dx} = 4x - 2xy

The right side is a difference of two terms, so you can’t split it into an

x

part times a

y

part yet. For separation of variables to work,

\frac{dy}{dx}

has to equal a product: a function of

x

multiplied by a function of

y

The Right Way

Rearrange:

\frac{dy}{dx} = 4x - 2xy

Factor out $2x$ :

\frac{dy}{dx} = 2x(2 - y)

Now it separates:

\frac{dy}{2 - y} = 2x \, dx

Integrate:

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

-\ln|2 - y| = x^2 + C

\ln|2 - y| = -x^2 + C

Exponentiate:

|2 - y| = e^{-x^2 + C} = Ce^{-x^2}

2 - y = Ce^{-x^2}

y = 2 - Ce^{-x^2}

Answer:

y = 2 - Ce^{-x^2}

Key Technique Summary

Situation	Action
$\frac{dy}{dx}$ with separate $x$ and $y$ functions	Separate and integrate
Result has $\ln\\|y\\|$	Exponentiate both sides
After exponentiating: $e^{f(x) + C}$	Rewrite as $Ce^{f(x)}$
Need an explicit solution	Solve algebraically for $y$
Even root (square root, 4th root, etc.)	Include $\pm$
Odd root (cube root, 5th root, etc.)	No $\pm$ needed

Strategy Flowchart

Step 1: Can you factor to get

\frac{dy}{dx}

multiplied by a function?

↓ Yes

Step 2: Separate the variables:

g(y) \, dy = f(x) \, dx

↓

Step 3: Integrate both sides. One

C

on the right side is enough

↓

Step 4: If you have

\ln|y|

, exponentiate both sides

↓

Step 5: Simplify:

e^{f(x) + C} = Ce^{f(x)}

↓

Step 6: Solve for

y

explicitly if you can

↓

Step 7: Check your answer by differentiating

Key Takeaways

Separation of variables: rearrange to get all the $y$ terms with $dy$ and all the $x$ terms with $dx$

Not every equation separates. $\frac{dy}{dx}$ has to be a function of $x$ times a function of $y$ , so something like $\frac{dy}{dx} = x + y$ can’t be separated

Leibniz notation: treat $\frac{dy}{dx}$ as if it were a fraction for separation purposes

General solutions contain the arbitrary constant $C$

One constant is enough. Write $C$ on only one side after integrating

Factor first. If $\frac{dy}{dx}$ is added to other terms, factor to get it into multiplication form

Natural log solutions: $\ln|y| = f(x) + C$ becomes $y = Ce^{f(x)}$ after exponentiating

Exponent rule: $e^{f(x) + C} = e^{f(x)} \cdot e^C = Ce^{f(x)}$ , so the constant moves outside

Even roots need $\pm$ . When solving $y^2 = \ldots$ , write $y = \pm\sqrt{\ldots}$

Odd roots don’t need $\pm$ . When solving $y^3 = \ldots$ , write $y = \sqrt[3]{\ldots}$

Back Next

7.6 - Finding General Solutions Using Separation of Variables

Review: Implicit Differentiation

General vs. Particular Solutions

What is Separation of Variables?

Leibniz Notation

The Separation of Variables Process

Example 1: Basic Separation

Implicit vs. Explicit Solutions

Implicit Solution

Explicit Solution

Example 2: Trigonometric Differential Equation

Example 3: Same Idea, Different Wording

Example 4: Negative Exponent Form

Example 5: The Reciprocal Technique

Example 6: Natural Logarithm and Exponentiating

Why We Rewrite ex2/2+Ce^{x^2/2 + C}ex2/2+C as Cex2/2Ce^{x^2/2}Cex2/2

Example 7: Another Exponential Solution

Example 8: Factoring Before Separating

The Wrong Way

The Right Way

Key Technique Summary

Strategy Flowchart

Key Takeaways

Why We Rewrite $e^{x^2/2 + C}$ as $Ce^{x^2/2}$