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PAGE 1

Separable Diff. Eqs.

Let's revisit:

\[ \frac{dy}{dx} = xy \]

Solvable by calculus:

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

Integrate:

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

Substitution Details

\[ u = y, \quad du = \frac{dy}{dx} \, dx \] \[ \int \frac{1}{u} \, du = \int x \, dx \]

Explanation: Applying standard \( u \)-substitution with \( u = y(x) \) transforms the integral with respect to \( x \) into an integral with respect to \( u \) via the Chain Rule.

But since \( u = y \), therefore \( du = dy \), the eq. above is really:

\[ \int \frac{1}{y} \, dy = \int x \, dx \]
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Separable Diff. Eqs. (Shortcut & Justification)

It looked as if we divided by \( y \) and “multiplied by \( dx \)”:

\[ \text{from } \frac{dy}{dx} = xy \longrightarrow \frac{1}{y} \, dy = x \, dx \]

↑ “multiplying by \( dx \)” is formally illegal

Then integrate.

\( \frac{dy}{dx} \) is NOT a quotient, it's a limit, so we cannot split up the “numerator” and “denominator”.

Even though multiplying by \( dx \) is NOT allowed, the end result is correct (justified by chain rule).

Key Takeaway: While Leibniz notation \( \frac{dy}{dx} \) is not an algebraic fraction, treating differentials as separable entities provides a valid mnemonic shortcut because the substitution rule / Chain Rule guarantees mathematical equivalence.

→ For the purpose of solving diff. eqs., we will use this as a shortcut.

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Separable Differential Equations

separable diff. eqs. are of the form

\[ \frac{dy}{dx} = \frac{g(x)}{f(y)} \]
A differential equation is separable if it can be factored such that all terms involving \( y \) can be grouped with \( dy \) and all terms involving \( x \) can be grouped with \( dx \).

such that we can separate the variables

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

then solve by integrating both sides

Example

for example,

\[ \frac{dy}{dx} = \frac{\sin(x)}{y} \]
  • separate variables: \( y \, dy = \sin(x) \, dx \)
  • integrate: \( \int y \, dy = \int \sin(x) \, dx \)
PAGE 4

Evaluating Integrals & Forms of Solutions

\[ \frac{1}{2} y^2 = -\cos(x) + \underbrace{C}_{\substack{\text{accounts for BOTH} \\ \text{constants of integration}}} \]

rewrite:

\[ y^2 + 2\cos(x) = C \]

really \( 2 \cdot \text{previous } C \) but \( C \) is constant so \( 2C \) is constant, we call it \( C \) again

the equation above is the solution (general) in the implicit form (\( y(x) \) is not explicitly stated)

Sometimes explicit solution is possible

for example,

\[ \begin{aligned} y^2 + 2\cos(x) &= C \\[6pt] y^2 &= C - 2\cos(x) \end{aligned} \]
\[ y(x) = \pm \sqrt{C - 2\cos(x)} \]
Taking the square root introduces both positive and negative roots (\( \pm \)). In initial value problems (IVPs), the appropriate branch is selected to satisfy the initial condition \( y(x_0) = y_0 \).
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Resolving Sign Ambiguity

How can that sign ambiguity be resolved?

We need one point on \( y(x) \) (e.g. initial condition).

For example, if \( y(0) = 1 \):

\[ y(0) = 1 = \pm \sqrt{C - 2\cos(0)} \]
\[ \underbrace{1}_{\text{positive}} = \pm \underbrace{\sqrt{C - 2}}_{\text{positive}} \]

Square Root Convention:

\( \sqrt{\quad} \rightarrow \text{always the positive solution} \) (the radical symbol denotes the principal non-negative square root).

So, we must choose \( + \):

\[ 1 = \sqrt{C - 2} \quad \text{so} \quad C = 3 \]

So, the particular solution is:

\[ y(x) = \sqrt{3 - 2\cos(x)} \]
PAGE 6

Solution Curves for \( \frac{dy}{dx} = \frac{\sin(x)}{y} \)

Visual representation of the implicit family of solutions \( y^2 + 2\cos(x) = C \) across different values of \( C \):

Coordinate plot showing slope field and solution curves of \frac{dy}{dx} = \frac{\sin(x)}{y} for values C = 0, 1.0, 2.0, 3.0, 5.0, 8.0 .
Figure: Coordinate plot showing slope field and solution curves of \( \frac{dy}{dx} = \frac{\sin(x)}{y} \) for values \( C = 0, 1.0, 2.0, 3.0, 5.0, 8.0 \). Show Details
Phase portrait and slope field for the differential equation \( \frac{dy}{dx} = \frac{\sin(x)}{y} \) corresponding to the level curves \( y^2 + 2\cos(x) = C \). The horizontal axis represents \( x \in [-3\pi, 3\pi] \) and the vertical axis represents \( y \in [-4, 4] \). Solutions are categorized by the parameter \( C \): closed orbits around centers \( (\pm \pi, 0) \) for \( C = 0 \) (blue) and \( C = 1.0 \) (yellow-green); the separatrix through saddles \( (2k\pi, 0) \) for \( C = 2.0 \) (green); and unbounded undulating horizontal bands for \( C = 3.0 \) (red), \( C = 5.0 \) (purple), and \( C = 8.0 \) (brown).

Phase Portrait Observations:

  • Center equilibria at \( x = \pm \pi, \pm 3\pi, \dots \), \( y = 0 \): Trajectories for \( 0 \le C < 2 \) form closed, bounded periodic orbits (concentric ovals) around stable centers.
  • Saddle points at \( x = 0, \pm 2\pi, \dots \), \( y = 0 \): The separatrix at \( C = 2.0 \) crosses through saddle equilibrium points where \( y = 0 \).
  • Continuous wavy trajectories for \( C > 2 \): Solution branches for \( C = 3.0, 5.0, 8.0 \) stay strictly positive (upper half-plane) or strictly negative (lower half-plane), undulating periodically without crossing \( y = 0 \).
PAGE 7

Interval of Validity for Differential Equation Solutions

Solution Family & First Initial Condition

\[ y^2 + 2\cos(x) = C \]

If \( y(0) = 1 \):

\[ y(x) = \sqrt{3 - 2\cos(x)} \]

Looking at this by itself, it appears all possible \( x \) is allowed: \( -\infty < x < \infty \) (confirmed by looking at graph)

Why all \( x \) is valid for \( C = 3 \):

Since \( -1 \le \cos(x) \le 1 \), we have \( 1 \le 3 - 2\cos(x) \le 5 \). The quantity under the square root is strictly positive everywhere, so \( y(x) \) and its derivative are defined on \( (-\infty, \infty) \).

Second Initial Condition & True Interval of Validity

Suppose a different initial condition led to \( C = 1 \):

\[ y(x) = \sqrt{1 - 2\cos(x)} \]

Look at this alone, domain is \( \frac{\pi}{3} \le x \le \frac{5\pi}{3} \).

BUT, this is NOT the true interval of validity.

  • \( \rightarrow \) Solution must be defined and differentiable AND satisfying the differential equation.

Endpoint Singularity:

At \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \), \( y(x) = 0 \). At these points, the derivative \( \frac{dy}{dx} = \frac{\sin(x)}{y} \) is undefined due to division by zero. Thus, the solution is only valid on the open interval \( \left(\frac{\pi}{3}, \frac{5\pi}{3}\right) \).

PAGE 8

Finding the Interval of Validity

\[ \frac{dy}{dx} = \frac{\sin(x)}{y} \]

Notice \( y\left(\frac{\pi}{3}\right) = y\left(\frac{5\pi}{3}\right) = 0 \), which causes a problem in the diff. eq.

\( \rightarrow \) The tangent line is vertical.

Interval of Validity Rule

To find the interval of validity of the solution, we need to find the interval on which both \( y \) and \( \frac{dy}{dx} \) exist and containing the initial condition.

Summary of Validity Requirements:

  • \( y(x) \) must be real and continuous.
  • \( y'(x) \) must exist (differentiable everywhere on the open interval).
  • The differential equation must be satisfied at every point in the interval.
  • The interval must be connected (an unbroken interval containing the initial point \( x_0 \)).

Integrating Factor as a Separable Differential Equation

We could have treated the diff. eq. for the integrating factor as a separable diff. eq:

\[ \frac{d\mu}{dt} = \mu \, p(t) \]

Separation of Variables Derivation:

\[ \frac{1}{\mu} d\mu = p(t)\,dt \implies \ln|\mu| = \int p(t)\,dt \implies \mu(t) = e^{\int p(t)\,dt} \]

PAGE 9

“multiply by \( dt \)”   divide by \( \mu \)

\[ \frac{1}{\mu} \, d\mu = p(t) \, dt \]
\[ \int \frac{1}{\mu} \, d\mu = \int p(t) \, dt \quad \text{and so on} \]

Concept Note: Integrating both sides yields \( \ln |\mu| = \int p(t) \, dt \), leading directly to the standard integrating factor formula \( \mu(t) = e^{\int p(t) \, dt} \).

Equations Both Linear and Separable

Some eqs. are both linear and separable

for example,

\[ \frac{dy}{dt} = ky \implies y' \underbrace{- k}_{p(t)} y = \underbrace{0}_{g(t)} \]

Explanation: \( \frac{dy}{dt} = ky \) can be solved either by separation of variables (\( \frac{1}{y} \, dy = k \, dt \)) or as a linear first-order equation \( y' + p(t)y = g(t) \) with \( p(t) = -k \) and \( g(t) = 0 \).

Non-Separable Linear Equations

\[ \text{non-separable} \qquad \frac{dy}{dx} = x + y \quad \text{(linear)} \]

subtract \( y \), “multiply by \( dx \)”

\[ \frac{dy}{dx} - y = x \] \[ dy - \underbrace{y \, dx}_{?} = x \, dx \]

Why Separation Fails: Writing the equation as \( dy = (x + y) \, dx \) shows that the terms \( x \) and \( y \) are summed and cannot be factored into a product of the form \( f(x)g(y) \). Thus, the differential equation is linear but non-separable.