Separable Equations

A separable equation is the friendliest kind of differential equation you will ever meet. It is the one type where, with a little algebraic rearranging, you can pull all the $x$'s to one side and all the $y$'s to the other, integrate each side on its own, and read off the answer. If differential equations feel intimidating, this is the doorway in: almost every idea in the subject — modeling growth, decay, cooling, mixing, and dilution — can be introduced through separable equations because the technique is transparent and the payoff is immediate.

The reason separable equations matter so much is that many of nature's most important laws say exactly the same thing: the rate at which something changes is proportional to how much of it there is right now. That single sentence, written as $\frac{dy}{dt} = ky$, is separable, and its solution — the exponential function — governs radioactive decay, compound interest, population booms, and drug clearance in the bloodstream. Learn to solve separable equations well and you have learned to model an enormous slice of the real world.

Learning Objectives

• Recognize when a first-order ODE is separable and rewrite it in the form $g(y)\,dy = f(x)\,dx$.
• Solve separable equations by integrating both sides and including the constant of integration correctly.
• Derive and apply the exponential growth and decay model $y = y_0 e^{kt}$.
• Solve initial-value problems (IVPs) by using a given condition to pin down the arbitrary constant.
• Interpret solutions physically — half-life, doubling time, and long-run behavior.
• Avoid the classic algebra and calculus pitfalls that make "separable" problems go wrong.

Quick Answer

A first-order differential equation is separable if you can write it as $\frac{dy}{dx} = f(x)\,g(y)$ — a product of a function of $x$ alone and a function of $y$ alone. To solve it, divide by $g(y)$ and multiply by $dx$ so that everything with $y$ sits on the left and everything with $x$ sits on the right: $\frac{1}{g(y)}\,dy = f(x)\,dx$. Then integrate both sides, adding a single constant $C$. The most famous case, $\frac{dy}{dt} = ky$, separates to $\frac{dy}{y} = k\,dt$ and integrates to $y = y_0 e^{kt}$ — exponential growth when $k > 0$ and decay when $k < 0$. If you are also given an initial condition like $y(0) = y_0$, substitute it to solve for $C$ and obtain the one specific solution.

Where It Came From

When Isaac Newton and Gottfried Wilhelm Leibniz invented calculus in the late 1600s, they immediately faced a new kind of problem: equations that related a quantity to its own rate of change. These differential equations described falling bodies, orbiting planets, and vibrating strings — but there was no general recipe for solving them, and there still isn't one today.

Separable equations were the very first class that mathematicians could crack systematically, and that is precisely why they are historically important. Leibniz himself, in correspondence around 1691, described the method of "separation of variables" — the intuitive act of gathering like differentials together. His notation, the $\frac{dy}{dx}$ that looks like a genuine fraction, was not an accident: Leibniz designed it so that the differentials $dy$ and $dx$ could be manipulated almost like ordinary numbers, making separation feel natural. This is one of the great advantages of Leibniz's notation over Newton's dot notation, and it is a major reason his notation is the one we still use.

The deeper motivation was necessity. Johann Bernoulli, Jacob Bernoulli, and later Leonhard Euler were trying to solve concrete problems — the shape of a hanging chain, the path of fastest descent, the growth of populations — and separable equations were the tools that actually yielded closed-form answers. Thomas Malthus's 1798 model of population growth and, much later, the 1900s understanding of radioactive decay both reduce to the same separable equation $\frac{dy}{dt} = ky$. The method survived three centuries not because it is elegant, but because it works and because so many real laws happen to be separable.

Concept 1: Recognizing and Separating

The defining test is simple. A first-order equation is separable if its right-hand side factors into a part depending only on $x$ times a part depending only on $y$:

$\frac{dy}{dx} = f(x)\,g(y).$

For example, $\frac{dy}{dx} = xy$ is separable ($f(x) = x$, $g(y) = y$). So is $\frac{dy}{dx} = \frac{x^2}{y}$, because it factors as $x^2 \cdot \frac{1}{y}$. But $\frac{dy}{dx} = x + y$ is not separable — a sum cannot be split into a clean product — and neither is $\frac{dy}{dx} = \sin(xy)$.

Once you see the product structure, the move is mechanical. Treat $\frac{dy}{dx}$ as a ratio of differentials (legitimate here as a bookkeeping device), divide both sides by $g(y)$, and multiply both sides by $dx$:

$\frac{1}{g(y)}\,dy = f(x)\,dx.$

Now each side involves only one variable, and you integrate.

Worked Example. Solve $\frac{dy}{dx} = xy$.

Separate: $\frac{1}{y}\,dy = x\,dx$.

Integrate both sides:

$\int \frac{1}{y}\,dy = \int x\,dx \quad\Longrightarrow\quad \ln|y| = \frac{x^2}{2} + C.$

Solve for $y$ by exponentiating:

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

Since $e^C$ is just a positive constant and $y$ may be positive or negative, we absorb everything into a single constant $A$ (which may be any real number, including zero for the trivial solution $y = 0$):

$y = A\,e^{x^2/2}.$

Notice we needed only one constant even though we integrated two sides — the two constants of integration combine into one. That single arbitrary constant is the signature of a first-order equation.

Concept 2: Exponential Growth and Decay

The single most important separable equation is

$\frac{dy}{dt} = ky,$

which says the rate of change of $y$ is proportional to $y$ itself. Separate and integrate:

$\int \frac{1}{y}\,dy = \int k\,dt \quad\Longrightarrow\quad \ln|y| = kt + C \quad\Longrightarrow\quad y = A\,e^{kt}.$

If $y = y_0$ when $t = 0$, then $A = y_0$, giving the clean model

$y = y_0\,e^{kt}.$

When $k > 0$ the quantity grows exponentially; when $k < 0$ it decays.

Worked Example — Radioactive Decay. A sample of a radioactive isotope decays at a rate proportional to the amount present, with $k = -0.05$ per year. If we start with 200 grams, how much remains after 30 years, and what is the half-life?

The model is $y = 200\,e^{-0.05t}$. After 30 years:

$y = 200\,e^{-0.05 \times 30} = 200\,e^{-1.5} \approx 200 \times 0.2231 \approx 44.6 \text{ grams}.$

For the half-life, we want the time $t_{1/2}$ when $y = 100$ (half of 200):

$100 = 200\,e^{-0.05 t_{1/2}} \;\Longrightarrow\; 0.5 = e^{-0.05 t_{1/2}} \;\Longrightarrow\; \ln 0.5 = -0.05\,t_{1/2}.$

So $t_{1/2} = \frac{\ln 0.5}{-0.05} = \frac{-0.6931}{-0.05} \approx 13.9$ years. Notice the half-life does not depend on the starting amount — a hallmark of exponential decay.

Concept 3: Initial-Value Problems

A separable equation on its own has infinitely many solutions, one for each value of the constant $C$ — a whole family of curves. An initial-value problem (IVP) adds a condition such as $y(0) = 3$ that selects exactly one member of that family. The strategy is always: (1) find the general solution with its constant, then (2) plug in the initial condition and solve for the constant.

Worked Example. Solve the IVP $\frac{dy}{dx} = \frac{6x^2}{y}$ with $y(0) = 4$.

Separate: $y\,dy = 6x^2\,dx$. Integrate:

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

Apply $y(0) = 4$: $\frac{16}{2} = 0 + C$, so $C = 8$. Then $\frac{y^2}{2} = 2x^3 + 8$, i.e. $y^2 = 4x^3 + 16$. Because $y(0) = 4$ is positive, we take the positive root:

$y = \sqrt{4x^3 + 16}.$

Check: at $x = 0$, $y = \sqrt{16} = 4$. ✓ And differentiating implicitly, $2y\,\frac{dy}{dx} = 12x^2$, so $\frac{dy}{dx} = \frac{6x^2}{y}$. ✓

Worked Example — Newton's Law of Cooling. A cup of coffee at $90^\circ$C sits in a $20^\circ$C room. Its temperature obeys $\frac{dT}{dt} = -k(T - 20)$. After 5 minutes it has cooled to $70^\circ$C. Find $T(t)$.

Let $u = T - 20$, so $\frac{du}{dt} = -ku$, which we already know gives $u = u_0 e^{-kt}$. At $t = 0$, $u_0 = 90 - 20 = 70$, so $T - 20 = 70\,e^{-kt}$. At $t = 5$, $T = 70$, so $50 = 70\,e^{-5k}$, giving $e^{-5k} = \frac{5}{7}$ and $k = \frac{1}{5}\ln\frac{7}{5} \approx 0.0673$ per minute. Thus

$T(t) = 20 + 70\,e^{-0.0673 t}.$

As $t \to \infty$, $T \to 20^\circ$C — the coffee approaches room temperature, exactly as physical intuition demands.

Real-World Applications

• Radiometric dating. Carbon-14 decays exponentially with a half-life of about 5730 years; archaeologists invert $y = y_0 e^{kt}$ to date bones and artifacts.
• Pharmacokinetics. Many drugs leave the bloodstream at a rate proportional to their concentration, so dosing schedules are computed from $C = C_0 e^{-kt}$.
• Population and epidemiology. Early, unconstrained population or infection growth is exponential; the separable logistic equation later corrects it for limited resources.
• Finance. Continuously compounded interest satisfies $\frac{dP}{dt} = rP$, giving $P = P_0 e^{rt}$ — the same equation as unchecked growth.
• Engineering — RC circuits. The voltage across a discharging capacitor obeys a separable equation and decays exponentially with time constant $RC$.

Common Mistakes

Mistake 1: Forgetting the constant of integration. Students integrate both sides but drop $+C$. Without it you get one particular curve instead of the general family, and you cannot satisfy an initial condition. Correction: always add exactly one constant, conventionally on the $x$-side after integrating.

Mistake 2: Trying to separate a non-separable equation. Faced with $\frac{dy}{dx} = x + y$, students write $\frac{dy}{y} = (x)\,dx$ or similar nonsense. A sum of $x$ and $y$ terms cannot be separated. Correction: confirm the right side factors as $f(x)g(y)$ before separating; if it does not, you need a different method (e.g. an integrating factor for linear equations).

Mistake 3: Mishandling the absolute value and the constant when exponentiating. From $\ln|y| = kt + C$, students sometimes write $y = e^{kt} + e^C$. But $e^{a+b} = e^a e^b$, not $e^a + e^b$. Correction: $y = e^{kt+C} = e^C e^{kt} = A e^{kt}$; the constant multiplies, it does not add. Also remember to plug in the initial condition to find $A$, not $C$.

Comparison and Connections

Separable equations are one family within first-order ODEs. It helps to see where they sit relative to their close cousins:

Type	Standard form	Solved by
Separable	$\dfrac{dy}{dx} = f(x)\,g(y)$	Separate variables, integrate
Linear first-order	$\dfrac{dy}{dx} + P(x)y = Q(x)$	Integrating factor $e^{\int P\,dx}$
Exact	$M\,dx + N\,dy = 0$ with $M_y = N_x$	Find potential function
Autonomous	$\dfrac{dy}{dt} = g(y)$	Separable special case; analyze equilibria

Note the overlap: an autonomous equation ($f(x) = 1$) is automatically separable, and the growth equation $\frac{dy}{dt} = ky$ is simultaneously separable, linear, and autonomous — which is why it is the perfect first example. When an equation is both separable and linear, separation is usually the quicker route.

Practice Questions

Recall

State the general form of a separable first-order differential equation and describe in one sentence the two steps used to solve it.

Answer: The form is $\frac{dy}{dx} = f(x)\,g(y)$. Step 1: rearrange to $\frac{1}{g(y)}\,dy = f(x)\,dx$. Step 2: integrate both sides, adding a single constant.

Understanding

Explain why the equation $\frac{dy}{dx} = x + y$ is not separable, while $\frac{dy}{dx} = xy$ is.

Answer: Separability requires the right side to be a product of a pure-$x$ function and a pure-$y$ function. In $xy$ the variables are already multiplied, so dividing by $y$ isolates each variable. In $x + y$ the variables are added; there is no algebraic way to move all $y$ terms to one side while keeping $x$ on the other, so the method fails.

Application

Solve $\frac{dy}{dx} = \frac{\cos x}{y}$ with $y(0) = 2$.

Answer: Separate: $y\,dy = \cos x\,dx$. Integrate: $\frac{y^2}{2} = \sin x + C$. Apply $y(0)=2$: $\frac{4}{2} = 0 + C$, so $C = 2$. Then $y^2 = 2\sin x + 4$, and taking the positive root (since $y(0) = 2 > 0$), $y = \sqrt{2\sin x + 4}$.

Analysis

A bacterial culture grows so that it doubles every 3 hours. Write its growth law and find how long it takes to reach ten times its initial size.

Answer: Growth is exponential: $y = y_0 e^{kt}$. Doubling in 3 hours means $2 = e^{3k}$, so $k = \frac{\ln 2}{3} \approx 0.231$ per hour. To reach $10 y_0$: $10 = e^{kt} \Rightarrow t = \frac{\ln 10}{k} = \frac{\ln 10}{(\ln 2)/3} = 3\cdot\frac{\ln 10}{\ln 2} \approx 3 \times 3.322 \approx 9.97$ hours.

FAQ

Is $\frac{dy}{dx}$ really a fraction I can split apart? Strictly, $\frac{dy}{dx}$ is a limit, not a fraction. But separation of variables is a legitimate shortcut that can be fully justified by the chain rule and integration; the differential manipulation always gives the correct answer for separable equations, which is exactly why Leibniz's notation is designed to look fraction-like.

What happened to the second constant of integration? You do get one from each side, but $C_1$ and $C_2$ combine: moving both to one side leaves a single arbitrary constant $C = C_2 - C_1$. A first-order equation always has just one free constant.

Why do I sometimes get $y = 0$ as a "lost" solution? When you divide by $g(y)$, you assume $g(y) \neq 0$. If $g(y_0) = 0$ for some constant $y_0$, then $y = y_0$ is also a solution (an equilibrium) that division hides. Always check these separately.

Do I need the absolute value in $\ln|y|$? Yes, because $\int \frac{1}{y}\,dy = \ln|y| + C$ for $y \neq 0$. In practice, once you exponentiate and fold the sign into the arbitrary constant $A$, the absolute value disappears and $A$ is allowed to be negative.

How do I know whether to keep the positive or negative square root? Use the initial condition. If solving gives $y^2 = \text{(something)}$ and you know $y$ is positive at the initial point, take the positive root; the solution stays on that branch as long as it remains continuous.

Quick Revision

• Test for separable: right side factors as $f(x)\,g(y)$.
• Method: $\frac{1}{g(y)}\,dy = f(x)\,dx$, then integrate both sides, add one constant $C$.
• Growth/decay: $\frac{dy}{dt} = ky \Rightarrow y = y_0 e^{kt}$; $k>0$ grows, $k<0$ decays.
• Half-life: $t_{1/2} = \frac{\ln 2}{|k|}$; doubling time: $t_2 = \frac{\ln 2}{k}$.
• IVP: find general solution first, then substitute the condition to find the constant.
• Watch out: don't drop $+C$; don't add exponentials ($e^{a+b}=e^a e^b$); check for lost equilibrium solutions where $g(y)=0$.

Related Topics

Prerequisites

• Differential Equations overview
• Integration techniques (Calculus) — you must be comfortable with $\int \frac{1}{y}\,dy$ and $\int x^n\,dx$.
• Logarithms and exponentials from Algebra.

Related Topics

• Linear first-order equations and the integrating factor method.
• Exact equations and autonomous equations.

Next Topics

• The logistic equation (a separable model of constrained growth).
• Second-order linear differential equations.