Separable ODEs — technique, implicit solutions: Worked examples

If any word below is unfamiliar, it was built in the parent note the parent topic. We reuse, never re-contradict, that vocabulary.

The scenario matrix

Each row is a case class — a genuinely different situation the technique can throw at you. The last column says which worked example covers it.

#	Case class	What's different about it	Covered by
C1	Clean explicit + IVP	Separates, integrates, isolate $y$ , fix $C$	Ex 1
C2	Implicit-only answer	Cannot isolate $y$ cleanly — leave as $F (x, y) = C$	Ex 2
C3	Partial-fractions needed	$h (y)$ is a product $\Rightarrow$ split before integrating	Ex 3
C4	Equilibrium / lost solution	$h (y_{0}) = 0$ deletes a constant solution	Ex 3 & Ex 4
C5	Sign / absolute-value branch	$ln ∣ y ∣$ and which branch the IVP selects	Ex 4
C6	Degenerate: $g (x) = 0$	RHS is $0$ $\Rightarrow$ $y$ is constant	Ex 5
C7	Limiting / long-run behaviour	What happens as $x \to \infty$ or near a blow-up	Ex 6
C8	Real-world word problem	Translate words → ODE → units check	Ex 7 (cooling)
C9	Exam twist: disguised separable	Looks like a sum/exponent but factors	Ex 8

We work them in an order that builds up.

Ex 1 — Case C1: clean explicit answer with an initial value

Worked example Statement

Solve $\frac{d y}{d x} = - 2 x y$ with $y (0) = 5$ .

Forecast: guess the shape. The rate $d y / d x$ is negative when $x > 0$ and $y > 0$ , and its size grows with $x$ — so from $y (0) = 5$ we expect $y$ to fall fast, a bell-ish decay. Hold that picture.

1. Separate. $\frac{d y}{y} = - 2 x d x$ . Why this step? The right-hand side already factors as $g (x) h (y) = (- 2 x) (y)$ , so all $y$ goes left, all $x$ right — the defining move of the method.

2. Integrate both sides. $ln ∣ y ∣ = - x^{2} + C$ . Why this step? $\int \frac{d y}{y} = ln ∣ y ∣$ and $\int - 2 x d x = - x^{2}$ . One constant $C$ (parent: merge constants).

3. Exponentiate. $∣ y ∣ = e^{C} e^{- x^{2}}$ , so $y = A e^{- x^{2}}$ with $A = \pm e^{C}$ . Why this step? Undo the $ln$ . The $\pm$ from the absolute value folds into one new nonzero constant $A$ .

4. Fix $C$ via $y (0) = 5$ . $5 = A e^{0} = A$ , so $A = 5$ . Why this step? An initial value pins down the one free constant. $y = 5 e^{- x^{2}}$

Verify: differentiate: $y^{'} = 5 \cdot (- 2 x) e^{- x^{2}} = - 2 x (5 e^{- x^{2}}) = - 2 x y$ . ✓ And $y (0) = 5$ . ✓ It's the decaying bell we forecast (see the Gaussian curve figure later reused).

Ex 2 — Case C2: the answer refuses to become explicit

Ex 3 — Cases C3 + C4: partial fractions and two hidden equilibria

Worked example Statement

Solve $\frac{d y}{d x} = y (y - 2)$ , then list all solutions.

Forecast: the RHS is zero at $y = 0$ and $y = 2$ — expect two flat equilibrium lines, and the moving solutions to be pushed away from or toward them. Guess whether $y = 2$ is a "wall" solutions run away from.

1. Check $h (y) = 0$ first. $h (y) = y (y - 2) = 0$ at $y = 0$ and $y = 2$ , so $y \equiv 0$ and $y \equiv 2$ are equilibrium solutions. Why this step? Parent's SISC rule: find the constants before dividing, because dividing erases them.

2. Separate (for $y \neq = 0, 2$ ). $\frac{d y}{y ( y - 2 )} = d x$ .

3. Partial fractions. $\frac{1}{y ( y - 2 )} = \frac{- 1/2}{y} + \frac{1/2}{y - 2}$ . Why this step? Each simple piece integrates to a log — see Partial Fractions. (Cover-up: at $y = 0$ , $\frac{1}{0 - 2} = - \frac{1}{2}$ ; at $y = 2$ , $\frac{1}{2} = \frac{1}{2}$ .)

4. Integrate. $- \frac{1}{2} ln ∣ y ∣ + \frac{1}{2} ln ∣ y - 2 ∣ = x + C$ , i.e. $\frac{1}{2} ln \frac{y - 2}{y} = x + C$ .

5. Solve for $y$ . Multiply by 2, exponentiate: $\frac{y - 2}{y} = B e^{2 x}$ ( $B = \pm e^{2 C}$ ). Rearranged, $y = \frac{2}{1 - B e ^{2 x}} plus y \equiv 0, y \equiv 2.$ Why report the equilibria? The formula can never output $0$ or $2$ , so they'd be lost without step 1.

Verify: take $B = 0$ : formula gives $y = 2$ — consistent with the equilibrium. Differentiate the general form (done in VERIFY) to recover $y^{'} = y (y - 2)$ . ✓ This is the logistic-flavour split from The Logistic Equation.

Ex 4 — Cases C4 + C5: the sign branch chosen by the initial value

Worked example Statement

Solve $\frac{d y}{d x} = \frac{x}{y}$ with (a) $y (0) = 3$ and (b) $y (0) = - 3$ .

Forecast: $y^{2} - x^{2}$ will be constant — the solution curves are hyperbolas. The sign of $y (0)$ decides the upper or lower branch. Guess which sign goes where before reading on.

1. Note the danger line. $h (y) = y$ , but here $h$ sits in a denominator, so $y = 0$ is excluded (rate blows up), not an equilibrium. Why this step? $1/ h (y)$ vs $h (y)$ behave differently at zeros — always look.

2. Separate. $y d y = x d x$ .

3. Integrate. $\frac{y ^{2}}{2} = \frac{x ^{2}}{2} + C_{0} \Rightarrow y^{2} - x^{2} = C$ (set $C = 2 C_{0}$ ).

4a. Apply $y (0) = 3$ . $9 - 0 = C \Rightarrow C = 9$ , so $y^{2} = x^{2} + 9$ . Since $y (0) = 3 > 0$ , choose the positive root: $y = + x^{2} + 9$ . Why this step? $\cdot$ has two branches; $y (0) > 0$ selects the upper sheet — the IVP resolves the sign.

4b. Apply $y (0) = - 3$ . $9 = C$ again, but now $y (0) = - 3 < 0$ selects $y = - x^{2} + 9$ . $y = + x^{2} + 9 (case a), y = - x^{2} + 9 (case b) .$

Verify: for (a), $y^{'} = \frac{x}{x ^{2} + 9} = \frac{x}{y}$ . ✓ At $x = 0$ , $y = 3$ . ✓ For (b) the same with an overall minus keeps $y^{'} = x / y$ . ✓ Look at the figure: the two branches never touch $y = 0$ , exactly the excluded line.

Ex 5 — Case C6: degenerate RHS, $g (x) = 0$

Ex 6 — Case C7: limiting / long-run and a finite-time blow-up

Worked example Statement

Solve $\frac{d y}{d x} = y^{2}$ with $y (0) = 1$ , then describe the behaviour as $x$ grows.

Forecast: the rate is the square of $y$ — feedback that accelerates itself. Guess: does $y$ stay finite forever, or explode at some finite $x$ ?

1. Check equilibrium. $h (y) = y^{2} = 0$ at $y = 0$ , so $y \equiv 0$ is an equilibrium solution. Why this step? SISC — record the lost constant.

2. Separate & integrate (for $y \neq = 0$ ). $\frac{d y}{y ^{2}} = d x \Rightarrow - \frac{1}{y} = x + C$ .

3. Solve for $y$ and fix $C$ . $y = \frac{- 1}{x + C}$ . At $x = 0$ , $y = 1 \Rightarrow - 1/ C = 1 \Rightarrow C = - 1$ . $y = \frac{1}{1 - x} .$

4. Limiting behaviour. As $x \to 1^{-}$ , denominator $\to 0^{+}$ , so $y \to + \infty$ : a finite-time blow-up at $x = 1$ . The solution only lives on $x < 1$ . Why this step? Squared feedback makes growth super-exponential — it escapes to infinity in finite $x$ , unlike $e^{x}$ .

Verify: $y^{'} = \frac{d}{d x} (1 - x)^{- 1} = (1 - x)^{- 2} = y^{2}$ . ✓ $y (0) = 1$ . ✓ The figure shows the vertical asymptote (plum dashed) at $x = 1$ .

Ex 7 — Case C8: real-world word problem (Newton's cooling), with units

Worked example Statement

A coffee at $9 0^{\circ} C$ sits in a $2 0^{\circ} C$ room. Its temperature $T (t)$ (in $^{\circ} C$ , $t$ in minutes) obeys Newton's law: the cooling rate is proportional to how far above room temperature it is, $\frac{d T}{d t} = - k (T - 20), k > 0.$ Given it cools to $7 0^{\circ} C$ after $10$ minutes, find $T (t)$ and the temperature at $t = 20$ .

Forecast: it should fall from $90$ toward $20$ and level off — an exponential approach to room temperature. Guess whether $t = 20$ is above or below $\frac{1}{2} (90 + 20) = 55$ .

1. Substitute to expose separability. Let $u = T - 20$ , so $\frac{d u}{d t} = \frac{d T}{d t} = - k u$ . Why this step? $u$ is the "excess over room" — the RHS becomes the clean factor $(- k) (u)$ .

2. Separate & integrate. $\frac{d u}{u} = - k d t \Rightarrow ln ∣ u ∣ = - k t + C \Rightarrow u = A e^{- k t}$ . Why $∣ u ∣$ ? Absolute value first; the $\pm$ folds into $A$ . Here $u > 0$ (coffee hotter than room), so $A > 0$ .

3. Back-substitute & use $T (0) = 90$ . $T = 20 + A e^{- k t}$ ; at $t = 0$ , $90 = 20 + A \Rightarrow A = 70$ .

4. Find $k$ from $T (10) = 70$ . $70 = 20 + 70 e^{- 10 k} \Rightarrow e^{- 10 k} = \frac{50}{70} = \frac{5}{7} \Rightarrow k = \frac{1}{10} ln \frac{7}{5}$ . $T (t) = 20 + 70 (\frac{5}{7})^{t /10} .$

5. Evaluate $t = 20$ . $T (20) = 20 + 70 (\frac{5}{7})^{2} = 20 + 70 \cdot \frac{25}{49} = 20 + \frac{250}{7} \approx 55.7 1^{\circ} C$ . Why this step? Answers the asked value.

Verify (units + sanity): $[k] = min^{- 1}$ , $[k t]$ dimensionless — the exponent is unitless. ✓ $T (0) = 90$ , $T (10) = 70$ , and $T \to 20$ as $t \to \infty$ (levels off at room temp). ✓ $55.71 > 55$ , just above the halfway mark — matches the slow-tail decay. ✓

Separable ODEs — Technique & Implicit Solutions (parent)
Initial Value Problems (fixing $C$ in Ex 1, 4, 5, 6, 7, 8)
Partial Fractions (Ex 3)
The Logistic Equation (Ex 3's cousin)
Exact ODEs (implicit solutions, Ex 2)
Substitution Method — Homogeneous ODEs (the $u = T - 20$ trick in Ex 7 is a baby substitution)
First-Order ODEs — Overview
Integrating Factor & Linear ODEs (when the RHS won't factor)

Worked examples — Separable ODEs — technique, implicit solutions

The scenario matrix

Ex 1 — Case C1: clean explicit answer with an initial value

Ex 2 — Case C2: the answer refuses to become explicit

Ex 3 — Cases C3 + C4: partial fractions and two hidden equilibria

Ex 4 — Cases C4 + C5: the sign branch chosen by the initial value

Ex 5 — Case C6: degenerate RHS, $g (x) = 0$

Ex 6 — Case C7: limiting / long-run and a finite-time blow-up

Ex 7 — Case C8: real-world word problem (Newton's cooling), with units

Ex 8 — Case C9: the exam twist — a disguised separable ODE

Did we fill every cell?

Connections