How do I know if a differential equation is separable?

A differential equation is separable if you can write it in the form \frac{dy}{dx} = g(x)h(y), where the right side is a product of a function depending only on x and a function depending only on y. If the right side is a sum or other combinations, the equation is not separable.

Do I always get a constant of integration on both sides when I integrate?

You can write a constant of integration on both sides (like \ln|y| = x^2 + C_1 and \ln|y| + C_2 = x^2), but it simplifies to keep just one constant C on the right side: \ln|y| = x^2 + C.

What if my answer is implicit, not explicit?

Many separable differential equations lead to implicit solutions (relations between x and y rather than y as a function of x). Both implicit and explicit solutions are valid. Use whichever form is more convenient for your problem or further analysis.

Why do I have to be careful with the constant of integration?

The constant of integration C represents the entire family of solutions to the differential equation. Forgetting it means you are missing infinitely many solutions, and an initial condition later will have no way to determine which solution fits your specific situation.

Separable Differential Equations

Solve differential equations where the rate of change is a product of a function of $x$ and a function of $y$ by separating variables and integrating.

\frac{dy}{dx} = g(x)h(y) \implies \int \frac{dy}{h(y)} = \int g(x)\,dx

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What each symbol means

What Separable Differential Equations takes

Symbol	Meaning
$g$	A function of $x$ only; appears as a factor in the differential equation. If you confuse this with $h(y)$ , you'll separate incorrectly and your integrals will be with respect to the wrong variable.
$h$	A function of $y$ only; appears as the other factor in the differential equation. Treating this as a function of $x$ will prevent you from separating the equation.
$x$	The independent variable; typically represents input values or (in word problems) time. Using $y$ as the independent variable reverses the role of everything in the method.
$y$	The dependent variable; the function you are solving for (expressed in terms of $x$ ). Swapping the roles of $x$ and $y$ will make the method not work.

When to use it

A differential equation fits this method when you can write it as $\frac{dy}{dx} = g(x)h(y)$ —a product of separate functions.

Level

Usually taught in: Calculus II · Appears on: AP Calculus

Worked examples

1. Solve $\frac{dy}{dx} = 2xy$

Problem

Solve the differential equation

\frac{dy}{dx} = 2xy

$\frac{1}{y} \, dy = 2x \, dx$
Separate variables by dividing both sides by $y$ and multiplying by $dx$ .
$\int \frac{1}{y} \, dy = \int 2x \, dx$
Set up integrals on both sides.
$\ln|y| = x^2 + C$
Integrate the left side using the standard result $\int \frac{1}{y} \, dy = \ln|y|$ and the right side using the power rule to get $\int 2x \, dx = x^2$ .
$y = Ae^{x^2}$
Exponentiate both sides; $A = \pm e^C$ is an arbitrary constant.

Answer: $y = Ae^{x^2}$

We can separate this equation because the right side is a product of a function of $x$ alone (which is $2x$ ) and a function of $y$ alone (which is $y$ ). Once separated, we integrate both sides and solve for $y$ .

2. Solve with negatives, fractions, and an initial condition

Problem

Solve

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

with the initial condition

y(0) = 2

$y \, dy = -3x \, dx$
Separate variables by multiplying both sides by $y$ and $dx$ .
$\int y \, dy = \int -3x \, dx$
Set up integrals on both sides.
$\frac{y^2}{2} = -\frac{3x^2}{2} + C$
Integrate using the power rule: $\int y \, dy = \frac{y^2}{2}$ and $\int -3x \, dx = -\frac{3x^2}{2}$ .
$y^2 + 3x^2 = 2C$
Multiply both sides by 2 and rearrange (treating $2C$ as a single constant).
$4 + 0 = 2C \implies 2C = 4$
Apply the initial condition $y(0) = 2$ to find the constant.

Answer: $y^2 + 3x^2 = 4$

This example requires care because $y$ appears in the denominator on the right side—we must move it to the left before integrating. Also, applying the initial condition $y(0) = 2$ finds the specific constant, giving us a particular solution instead of the general form.

3. Exponential growth: savings account

Problem

A savings account grows at a rate proportional to the amount of money in it. If the account currently has

\$1000

and grows at a rate of

5\%

per year, find

A(t)

, the amount after

t

years.

$\frac{dA}{dt} = 0.05A$
Set up the differential equation: the rate of change (left side) equals the proportionality constant (0.05, or 5% per year) times the current amount.
$\frac{1}{A} \, dA = 0.05 \, dt$
Separate variables by dividing both sides by $A$ .
$\ln A = 0.05t + C$
Integrate both sides (we can drop the absolute value since $A > 0$ ).
$A = Ke^{0.05t}$
Exponentiate both sides; $K = e^C$ is the arbitrary constant.
$K = 1000$
Apply the initial condition $A(0) = 1000$ to find that $1000 = Ke^0 = K$ .

Answer: $A(t) = 1000e^{0.05t}$

This real-world scenario leads to an exponential growth equation. Separable differential equations are essential for modeling compound growth because the differential equation naturally separates into a function of time alone and a function of the amount alone.

Common mistakes

Where Separable Differential Equations usually goes wrong

The mistake	Why it is wrong	The fix
Writing $\int \frac{dy}{h(y)} \, dx = \int g(x) \, dx$	You cannot integrate $\frac{dy}{h(y)}$ with respect to $x$ ; the left side has $dy$ and must be integrated with respect to $y$ .	Always match the variable of integration to the differential: $\int \frac{dy}{h(y)} = \int g(x) \, dx$ . The left side integrates over $y$ , the right over $x$ .
When solving $\frac{dy}{dx} = x^2 + y$ , separating as $\frac{dy}{x^2 + y} = dx$	This equation does not separate because $x^2 + y$ is not a product of a function of $x$ alone and a function of $y$ alone; the equation is additive, not multiplicative.	Separable equations must have the form $\frac{dy}{dx} = g(x)h(y)$ (a product). If you see a sum, you need a different method such as an integrating factor, which works for linear equations.
Finding $\int \frac{1}{y} \, dy = \frac{1}{y}$ instead of $\ln\|y\|$	This is a common integration error that breaks the entire solution; the integral of $\frac{1}{y}$ is the natural logarithm, not a power rule.	Remember that $\int \frac{1}{y} \, dy = \ln\|y\|$ , not $y^{-1}$ . The power rule $\int y^n \, dy = \frac{y^{n+1}}{n+1}$ does not apply when $n = -1$ ; the logarithm is the exception.

Tips and when to use something else

Before separating, check that the equation truly factors into a product of a function of $x$ and a function of $y$ . If it does not, the equation is not separable and you need a different method such as an integrating factor, which works for linear equations.
Always include the constant of integration $C$ on one side when you integrate; forgetting it will lose the entire family of solutions.
When you integrate $\int \frac{1}{y} \, dy$ , you get $\ln|y|$ , not $\frac{1}{y}$ . This is the only case where the power rule breaks down ( $n = -1$ ), so be extra careful.
After integrating, you may end up with an implicit solution (like $x^2 + y^2 = 4$ ) rather than an explicit one ( $y = \sqrt{4 - x^2}$ ). Both are valid; choose whichever is easier to work with.

Related formulas

Indefinite IntegralThe indefinite integral finds all functions whose derivative equals a given function, expressed as an antiderivative plus an arbitrary constant.Integral of 1/xIntegrating 1/x yields the natural logarithm of the absolute value of x, plus a constant—essential for reciprocal function problems.U-SubstitutionU-substitution simplifies integrals by transforming them into easier forms using a change of variables, ideal when you spot a function and its derivative.Integration by PartsIntegration by Parts is a calculus technique for evaluating integrals of products by transforming them into simpler forms when substitution fails.

Frequently asked questions

How do I know if a differential equation is separable?: A differential equation is separable if you can write it in the form $\frac{dy}{dx} = g(x)h(y)$ , where the right side is a product of a function depending only on $x$ and a function depending only on $y$ . If the right side is a sum or other combinations, the equation is not separable.
Do I always get a constant of integration on both sides when I integrate?: You can write a constant of integration on both sides (like $\ln|y| = x^2 + C_1$ and $\ln|y| + C_2 = x^2$ ), but it simplifies to keep just one constant $C$ on the right side: $\ln|y| = x^2 + C$ .
What if my answer is implicit, not explicit?: Many separable differential equations lead to implicit solutions (relations between $x$ and $y$ rather than $y$ as a function of $x$ ). Both implicit and explicit solutions are valid. Use whichever form is more convenient for your problem or further analysis.
Why do I have to be careful with the constant of integration?: The constant of integration $C$ represents the entire family of solutions to the differential equation. Forgetting it means you are missing infinitely many solutions, and an initial condition later will have no way to determine which solution fits your specific situation.

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Reviewed 2026-09-18