Separable Differential Equations
Quick Reference
| Field | Value |
|---|---|
| Textbook | Logan, A First Course in Differential Equations, 3rd edition |
| Chapter.Section.Subsection | 1.3.1 Separation of Variables |
| Pages | pp. 22-30 (Eq. (1.11); Remark 1.15) |
| Course | MATH347 |
The page references above are verified against the 3rd edition.
Try This First
Before any formula, work this small case with your own hands.
Suppose a quantity $x$ changes in time according to
\[ \frac{dx}{dt} = t\,x. \]
You do not yet have a method. You do have one honest move: ask which functions of $x$ belong on the left and which functions of $t$ belong on the right.
- Multiply both sides by $dt$ and divide both sides by $x$. Write down what lands on each side.
- One side now holds only $x$ and $dx$. The other holds only $t$ and $dt$. Notice that the variables have been pulled apart.
- Put an integral sign on each side. What two integrals do you get?
Check your separation
You should reach
\[ \frac{1}{x}\,dx = t\,dt, \qquad \int \frac{1}{x}\,dx = \int t\,dt. \]
The left integral is $\ln\lvert x\rvert$. The right integral is $\tfrac{1}{2}t^2$. Hold that thought. The only new idea is the move you just made.
A rule that ties the rate of change of $x$ to both $x$ and $t$ became two separate one-variable integrals. That separation is the technique you are learning here.
An Intuitive Introduction
A differential equation is a process statement. It does not hand you the function $x(t)$ directly. It tells you the slope of $x$ at every instant in terms of where you are ($x$) and what time it is ($t$). The job is to recover the function whose slope obeys that rule.
Some of these equations have a lucky structure. The right side factors cleanly into a part that depends only on $x$ times a part that depends only on $t$. When that happens, the two variables can be moved to opposite sides of the equation, and each side becomes an ordinary integral from second-semester calculus. The original equation collapses into two integration problems you already know how to do.
A direction field shows what solutions look like and a numerical method approximates one, but neither gives a formula. Separable equations are the first large family for which an exact formula is within reach, and the method to find it is direct.
The reason this works traces back to the chain rule and to $u$-substitution. Reading those rules in reverse is what lets you treat $dx$ and $dt$ as quantities you can move from one side to the other. The technique looks like algebra on symbols, and the rigorous justification is the substitution rule, so both views are correct.
There is one honest subtlety worth meeting up front. Integrating an $x$-side and a $t$-side usually produces an equation linking $x$ and $t$ that is not yet solved for $x$. That is called an implicit solution, and it is a complete answer even when you cannot untangle it into $x = (\text{something in } t)$.
Prerequisite Hub
Builds On
| Skill | Course | Why it is needed |
|---|---|---|
Introduction to differential equations (de-introduction) |
MATH347 | Defines what a first-order ODE and a solution are |
Antiderivatives and initial value problems (antiderivatives-ivp) |
MATH347 | Both sides get integrated; an initial condition fixes the constant |
Direction fields (direction-fields) |
MATH347 | A slope field gives a picture to check the answer against |
Chain rule and $u$-substitution (math347-w0-chain-rule-and-u-substitution) |
MATH347 | The separation step is the chain rule read backward |
Cross-Course Prerequisites
| Skill | Course | Why it is needed |
|---|---|---|
Antidifferentiation rules (antidifferentiation-rules) |
MATH162 | The standard antiderivative table powers both integrals |
Indefinite $u$-substitution (u-substitution-indefinite) |
MATH162 | The $t$-side and $x$-side integrals are substitution problems |
Initial value problems (initial-value-problems) |
MATH161 | An initial condition selects one curve from the family |
Implicit differentiation (implicit-differentiation) |
MATH161 | The solution often stays in implicit form before solving for $x$ |
Unlocks
| Skill | What it builds next |
|---|---|
First-order linear equations (first-order-linear) |
The integrating-factor method for equations that do not separate |
The logistic equation (logistic-equation) |
A separable population model with an equilibrium ceiling |
Applications of separable equations (separable-applications) |
Cooling, decay, mixing, and growth models |
Prerequisite Map
The Official Definition
Separable equation. A differential equation having the form $x' = f(x)g(t)$, where the right side is a product of a function of $x$ and a function of $t$, is called a separable equation.
Logan, 3rd ed., §1.3.1 Separation of Variables, p. 22, Eq. (1.11)
The test for separability is a factoring question. Look at the right side of $x' = \cdots$ and ask whether it splits into (a function of $x$ alone) times (a function of $t$ alone). If it does, the equation is separable. If the right side mixes the variables in a way that will not factor, such as $x + t$, then this method does not apply and a different technique is needed.
Method of separation of variables (recipe). For $\dfrac{dx}{dt} = f(x)g(t)$, the method of separation of variables results in writing down $\dfrac{1}{f(x)}\,dx = g(t)\,dt$, with the $x$ terms (including $dx$) on the left and the $t$ terms (including $dt$) on the right, and then integrating both sides: \[ > \int \frac{1}{f(x)}\,dx = \int g(t)\,dt. > \]
Logan, 3rd ed., Remark 1.15, p. 23
Equilibrium (constant) solutions and division by $f(x)$. Dividing by $f(x)$ to separate requires $f(x) \neq 0$; values of $x$ with $f(x) = 0$ give constant (equilibrium) solutions that must be checked separately because they are lost in the division step.
Logan, 3rd ed., §1.3.1, p. 22 (separation step; cf. equilibrium solutions p. 15, §1.5)
This third point is the one most readers skip, and it is the one that costs the most. The instant you divide by $f(x)$, you have quietly assumed $f(x)$ is never zero. Any constant value of $x$ that makes $f(x) = 0$ is a perfectly good solution, and the division step throws it away. Build the habit of finding those roots first, before you separate anything.
The Method, Step by Step
Goal: solve $\dfrac{dx}{dt} = f(x)g(t)$.
Step 1. Confirm the equation is separable. Check that the right side factors as a function of $x$ times a function of $t$.
Step 2. Find the equilibrium solutions. Solve $f(x) = 0$. Each root $x = c$ gives a constant solution $x(t) = c$. Set these aside; the next steps will not recover them.
Step 3. Separate. Move every $x$ (and the $dx$) to one side and every $t$ (and the $dt$) to the other:
\[ \frac{1}{f(x)}\,dx = g(t)\,dt. \]
Step 4. Integrate both sides. One constant of integration is enough; place a single $+C$ on the $t$-side.
Step 5. Apply the initial condition (if one is given) to solve for $C$.
Step 6. Solve for $x$ when possible. Many separable solutions are honest in implicit form, so do not force an explicit formula if the algebra fights back.
Step 7. State the full solution set: the explicit (or implicit) family plus any equilibrium solutions from Step 2.
Quick Reference
| Property | Value |
|---|---|
| Form | $\dfrac{dx}{dt} = f(x)g(t)$ |
| Separated | $\dfrac{1}{f(x)}\,dx = g(t)\,dt$, valid where $f(x) \neq 0$ |
| Integrated | $\displaystyle\int \frac{1}{f(x)}\,dx = \int g(t)\,dt + C$ |
| Watch for | constant solutions where $f(x) = 0$ (lost in the division step) |
| Answer form | explicit $x = (\text{expression in } t)$, or a valid implicit relation in $x$ and $t$ |
Worked Examples
Example 1: Finish the opener
Solve $\dfrac{dx}{dt} = t\,x$ with $x(0) = 5$.
Predict first. The right side is positive when $x$ and $t$ are both positive, so for $t > 0$ a positive starting value should grow. Expect an increasing function for $t > 0$, and expect something exponential in flavor because the rate is proportional to $x$ itself.
Step 1. The right side is $f(x)g(t) = x \cdot t$. Separable.
Step 2. $f(x) = x = 0$ at $x = 0$, so $x(t) = 0$ is an equilibrium solution. The initial value is $5$, not $0$, so the solution of interest is not the constant one, but record $x = 0$ as a solution of the equation.
Step 3. Separate:
\[ \frac{1}{x}\,dx = t\,dt. \]
Step 4. Integrate:
\[ \ln\lvert x\rvert = \frac{t^2}{2} + C. \]
Step 5. Exponentiate, then apply $x(0) = 5$:
\[ \lvert x\rvert = e^{C}\, e^{t^2/2}, \qquad x = A\, e^{t^2/2}, \]
where $A = \pm e^{C}$ is a new constant. At $t = 0$: $x(0) = A\,e^{0} = A = 5$.
Step 6. The solution is
\[ x(t) = 5\, e^{t^2/2}. \]
Check against the prediction. For $t > 0$ the exponent grows, so $x$ grows, matching the prediction. Verify by differentiating: $x'(t) = 5\, e^{t^2/2} \cdot t = t \cdot x(t)$, which is the original equation. The solution checks.
Example 2: An implicit answer
Solve $\dfrac{dx}{dt} = \dfrac{t}{x}$.
Predict first. The relationship looks symmetric in $t$ and $x$. A guess worth testing is that $x^2 - t^2$ stays constant. Hold that guess and see whether the method confirms it.
Step 1. Right side $= t \cdot \dfrac{1}{x}$, a function of $t$ times a function of $x$. Separable.
Step 2. $f(x) = \dfrac{1}{x}$ is never zero, so there are no equilibrium solutions. (The line $x = 0$ is excluded because the right side is undefined there, not because $f$ vanishes.)
Step 3. Separate:
\[ x\,dx = t\,dt. \]
Step 4. Integrate:
\[ \frac{x^2}{2} = \frac{t^2}{2} + C. \]
Step 5. No initial condition is given, so leave $C$ as a parameter. Multiply by $2$ and rename the constant:
\[ x^2 - t^2 = K. \]
Step 6. This is an implicit family of hyperbolas. Solving for $x$ gives $x = \pm\sqrt{t^2 + K}$, with the sign fixed by an initial condition. The implicit form $x^2 - t^2 = K$ is the cleaner answer.
Check against the prediction. The guess that $x^2 - t^2$ is constant was correct. Differentiate $x^2 - t^2 = K$ implicitly: $2x\,x' - 2t = 0$, so $x' = \dfrac{t}{x}$. The solution checks.
Example 3: An equilibrium solution that the recipe loses
Solve $\dfrac{dx}{dt} = x(2 - x)$ with $x(0) = 2$.
Predict first. The rate is zero exactly when $x = 0$ or $x = 2$. The starting value is $2$, which makes the rate zero. So the prediction is that $x$ never moves: $x(t) = 2$ for all $t$.
Step 1. Right side $= x(2 - x) \cdot 1$, a function of $x$ times a function of $t$ (the $t$-factor is $g(t) = 1$). Separable.
Step 2. $f(x) = x(2 - x) = 0$ at $x = 0$ and $x = 2$. So $x(t) = 0$ and $x(t) = 2$ are both equilibrium solutions.
Step 3 onward. The initial value $x(0) = 2$ is exactly an equilibrium. The constant function $x(t) = 2$ already satisfies the equation and the initial condition. Differentiating the constant gives $x'(t) = 0$, and the right side at $x = 2$ is $2(2 - 2) = 0$. Both sides equal zero, so $x(t) = 2$ solves it.
Why the recipe alone would miss this. Separating would require dividing by $x(2 - x)$, which is zero at $x = 2$. That division is illegal here, and the integral form would never produce the constant solution. This is precisely the case the third definition warns about: the equilibrium solution is outside the integrated family and has to be checked on its own.
Check against the prediction. The prediction (no motion) matches the equilibrium solution. The solution checks.
Common Misconceptions
the separation step is illegal symbol-pushing. Treating $\dfrac{dx}{dt}$ as a fraction and moving $dt$ across the equals sign looks like an abuse of notation. It is not magic, and it is not breaking a rule. Writing $\dfrac{1}{f(x)}\,dx = g(t)\,dt$ and integrating is shorthand for the substitution rule applied to $\displaystyle\int \frac{1}{f(x)}\frac{dx}{dt}\,dt = \int g(t)\,dt$. The mechanical move and the rigorous version give the same answer, so the shortcut is justified, not merely tolerated.
every constant of integration deserves its own $+C$. A reader who writes $\ln\lvert x\rvert + C_1 = \tfrac{1}{2}t^2 + C_2$ has not made an error, but the two constants collapse into one, because $C_2 - C_1$ is itself just a single unknown constant. Put one $+C$ on the right side and move on. Predict-then-check: try solving $\dfrac{dx}{dt} = x$ with two constants and then with one, and confirm that both lead to $x = A e^{t}$ with a single free parameter.
dividing by $f(x)$ is always safe. The division step assumes $f(x) \neq 0$. Consider $\dfrac{dx}{dt} = x^2 - x$ with $x(0) = 1$. The tempting move is to separate immediately. Predict-then-check: first solve $f(x) = x^2 - x = 0$, finding $x = 0$ and $x = 1$. Since $x(0) = 1$ is a root, the constant function $x(t) = 1$ is the solution, and any separated-and-integrated answer would have missed it entirely. Always hunt for the equilibrium solutions before dividing.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Treating a sum as separable | $\dfrac{dx}{dt} = x + t$ | Not separable; the right side must factor as a product |
| Dropping the absolute value | $\displaystyle\int \frac{1}{x}\, dx = \ln x$ | Keep $\ln\lvert x\rvert$ |
| Losing the equilibrium solution | dividing by $x$ drops $x = 0$ | Check separately where $f(x) = 0$ |
| Forgetting the constant of integration | omitting $+ C$ | Add $C$ before applying the initial condition |
| Refusing to leave an implicit answer | forcing $x = \cdots$ | An implicit relation in $x$ and $t$ is a valid solution |
| Applying the initial condition too early | substituting before integrating | Integrate to the general solution first, then find $C$ |
Leveled Practice
Level 1: Is It Separable?
Problem 1. For each equation, state whether the right side factors into a function of $x$ times a function of $t$.
(a) $\dfrac{dx}{dt} = 3t^2 x$ (b) $\dfrac{dx}{dt} = x + t$ (c) $\dfrac{dx}{dt} = \dfrac{\cos t}{x^2}$
Show answer
(a) Separable. The right side is $x \cdot 3t^2 = f(x)g(t)$ with $f(x) = x$ and $g(t) = 3t^2$.
(b) Not separable. The sum $x + t$ does not factor into a product of a function of $x$ and a function of $t$.
(c) Separable. The right side is $\dfrac{1}{x^2} \cdot \cos t$, with $f(x) = \dfrac{1}{x^2}$ and $g(t) = \cos t$.
Level 2: A Clean Separation
Problem 2. Solve $\dfrac{dx}{dt} = 3t^2 x$ with $x(0) = 4$.
Thought Process
The right side is $x \cdot 3t^2$. Check for equilibria ($f(x) = x = 0$), separate, integrate, then use $x(0) = 4$.
Show answer
Equilibrium: $x = 0$ (not the one we want, since $x(0) = 4$).
Separate and integrate:
\[ \frac{1}{x}\,dx = 3t^2\,dt, \qquad \ln\lvert x\rvert = t^3 + C. \]
Exponentiate: $x = A e^{t^3}$. Apply $x(0) = 4$: $A = 4$.
\[ x(t) = 4\, e^{t^3}. \]
Check: $x'(t) = 4 e^{t^3} \cdot 3t^2 = 3t^2 x(t)$. Correct.
Level 3: An Implicit Solution
Problem 3. Solve $\dfrac{dx}{dt} = \dfrac{t^2}{x}$ with $x(0) = 3$. Leave the answer in whatever form is cleanest.
Thought Process
Separate as $x\,dx = t^2\,dt$. Integrate both sides, then use the initial condition to pin the constant. Decide whether an explicit form is reasonable.
Show answer
No equilibria ($\dfrac{1}{x}$ is never zero). Separate and integrate:
\[ x\,dx = t^2\,dt, \qquad \frac{x^2}{2} = \frac{t^3}{3} + C. \]
Apply $x(0) = 3$: $\dfrac{9}{2} = 0 + C$, so $C = \dfrac{9}{2}$.
Implicit form: $\dfrac{x^2}{2} = \dfrac{t^3}{3} + \dfrac{9}{2}$, equivalently $x^2 = \dfrac{2t^3}{3} + 9$.
Explicit form (positive branch, since $x(0) = 3 > 0$): $x(t) = \sqrt{\dfrac{2t^3}{3} + 9}$.
Check: $2x\,x' = 2t^2$ gives $x' = \dfrac{t^2}{x}$. Correct.
Level 4: Catch the Equilibrium
Problem 4. Solve $\dfrac{dx}{dt} = x(x - 1)$ with $x(0) = 1$. State every solution of the equation that you encounter along the way.
Thought Process
Find the equilibria first by solving $f(x) = x(x-1) = 0$. Then notice whether the initial condition lands on one of them before doing any dividing.
Show answer
Equilibria: $x(x - 1) = 0$ at $x = 0$ and $x = 1$. So $x(t) = 0$ and $x(t) = 1$ are both solutions of the equation.
The initial value $x(0) = 1$ is exactly the equilibrium $x = 1$. The constant function $x(t) = 1$ satisfies the equation: $x'(t) = 0$ and the right side at $x = 1$ is $1(1 - 1) = 0$. Both sides are zero.
\[ x(t) = 1. \]
Separating would have required dividing by $x(x - 1)$, which is zero at $x = 1$, so the integrated family would have missed this answer. The equilibrium check is what saves the problem.
Level 5: Partial Fractions and the Logistic Pattern
Problem 5. Solve $\dfrac{dx}{dt} = x(1 - x)$ with $x(0) = \dfrac{1}{2}$. (Hint: $\dfrac{1}{x(1-x)} = \dfrac{1}{x} + \dfrac{1}{1-x}$.)
Thought Process
Record the equilibria $x = 0$ and $x = 1$. The starting value $\tfrac{1}{2}$ is between them, so the solution is not constant. Separate, integrate the left side with the partial-fraction hint, solve for $x$, and apply the initial condition.
Show answer
Equilibria: $x = 0$ and $x = 1$ (neither matches $x(0) = \tfrac{1}{2}$).
Separate: $\dfrac{1}{x(1 - x)}\,dx = dt$. Use the hint:
\[ \left(\frac{1}{x} + \frac{1}{1 - x}\right)dx = dt. \]
Integrate. The $\dfrac{1}{1-x}$ term integrates to $-\ln\lvert 1 - x\rvert$:
\[ \ln\lvert x\rvert - \ln\lvert 1 - x\rvert = t + C, \qquad \ln\left\lvert\frac{x}{1 - x}\right\rvert = t + C. \]
Exponentiate: $\dfrac{x}{1 - x} = A e^{t}$. Apply $x(0) = \tfrac{1}{2}$: $\dfrac{1/2}{1 - 1/2} = 1 = A$.
So $\dfrac{x}{1 - x} = e^{t}$. Solve for $x$:
\[ x = e^{t}(1 - x) \implies x(1 + e^{t}) = e^{t} \implies x(t) = \frac{e^{t}}{1 + e^{t}} = \frac{1}{1 + e^{-t}}. \]
This is the logistic curve. Check $x(0) = \dfrac{1}{1 + 1} = \dfrac{1}{2}$. As $t \to \infty$, $x \to 1$ (the upper equilibrium), and as $t \to -\infty$, $x \to 0$ (the lower equilibrium). The equilibria act as floor and ceiling for the rising solution, the pattern that the logistic equation generalizes.
Mastery Checklist
Novice (Level 1-2):
Competent (Level 3-4):
Proficient (Level 5):
Mental Model
Think of a separable equation as a knot that happens to come apart cleanly. The differential equation tangles $x$, $t$, and the rate $\dfrac{dx}{dt}$ together. The factoring condition $f(x)g(t)$ is the loose thread: pull on it and the $x$ pieces slide to one side while the $t$ pieces slide to the other. Once the knot is undone, each side is an ordinary integral.
The one piece that does not survive the pull is any value of $x$ where $f(x) = 0$. Those equilibrium values are the frozen strands, the constant solutions that sit still because their rate is zero. The integrating method cannot reach them, so they are found by inspection before any pulling begins.
Holding this view explains why separable equations are the gateway method. They are the case where the variables come apart cleanly, so a formula exists. The later methods (integrating factors and substitutions) are techniques for making a stubborn equation come apart when it does not separate on its own.
Connections
Looking back:
- Introduction to differential equations defines the objects this method solves
- Antiderivatives and initial value problems supplies the integration and the constant-fixing step
- Direction fields give a picture of the family of solutions to compare against
Looking ahead:
- First-order linear equations handle equations that do not separate, using an integrating factor
- The logistic equation is the separable model behind bounded population growth
- Applications of separable equations cover decay, cooling, and mixing problems
Real-world connections:
- Radioactive decay obeys $\dfrac{dx}{dt} = -kx$, separable with an exponential solution
- Newton’s law of cooling separates into a temperature-difference integral
- Population models with a carrying capacity separate into the logistic curve
Audience Notes
For students who find math intimidating: The method is a checklist. Factor the right side into an $x$-part and a $t$-part, move the $x$ terms (with $dx$) to one side and the $t$ terms (with $dt$) to the other, integrate each side, then use the starting value to find the constant. A wrong turn here is common and shows something worth understanding; there is no need to rush.
For students interested in proof: The legitimacy of treating $\dfrac{dx}{dt}$ as a ratio of differentials is the substitution-rule argument in the first misconception callout. The separation move is the chain rule run in reverse.
For students interested in careers: Separable equations model the cleanest real systems, including radioactive decay, simple mixing tanks, and first-order chemical reactions. They are the analytic baseline an engineer reaches for before resorting to numerical solvers for messier models.
Resources
| Resource | What it covers |
|---|---|
| Logan 3rd ed. §1.3.1 Separation of Variables (pp. 22-30; Eq. (1.11), Remark 1.15) | The primary text: definition, the separation recipe, and the worked treatment |
| OpenStax Calculus Vol. 2 §4.3 Separable Equations | An open-access second pass on the method and on lost (equilibrium) solutions |
| Logan 1.3 tutor guide (separable definition, step-by-step method, the absolute-value and log step) | Local companion guide at ~/math347-ingest/math_guides/MATH_347/Logan DEs/New Guides/Chapter 1/logan_1_3.tex |
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|---|---|---|
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Last updated: 2026-06-16