Separable Differential Equations: Solve Step By Step

What Are Separable Differential Equations?

Separable differential equations are first-order differential equations whose right-hand side factors into a part that depends only on $x$ and a part that depends only on $y$:

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

If you can write the equation in that product form, it is separable, and the whole solution method opens up. The idea is to gather every $y$ (with the $dy$) on one side and every $x$ (with the $dx$) on the other, then integrate each side by itself. That is why the technique is called separation of variables.

Two quick tests decide whether an equation qualifies:

• It is separable when the right side is a product of a function of $x$ alone and a function of $y$ alone, such as $\frac{dy}{dx} = xy$, $\frac{dy}{dx} = y^2$, or $\frac{dy}{dx} = \frac{x}{y}$.

• It is not separable when $x$ and $y$ are locked together in a way that will not factor, such as $\frac{dy}{dx} = x + y$. No rearrangement splits a sum like that into an $x$-only times a $y$-only piece.

Separable equations are the first solvable family in the study of ordinary differential equations, and they carry order and degree one and one: one derivative, raised to the first power.

How Do You Solve A Separable Differential Equation?

The method is four short moves. Start from $\frac{dy}{dx} = g(x),h(y)$ and assume for the moment that $h(y) \neq 0$.

1. Separate the variables. Divide by $h(y)$ and multiply by $dx$ so each variable sits with its own differential:

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

2. Integrate both sides. Integrate the left side with respect to $y$ and the right side with respect to $x$. Each side produces its own constant, but the two combine into a single constant $C$ on the right:

$$\int \frac{dy}{h(y)} = \int g(x),dx + C$$

3. Solve for $y$, or leave it implicit. If the resulting equation rearranges cleanly, write $y$ as an explicit function of $x$. If it does not, an implicit relation between $x$ and $y$ is a complete and acceptable answer.

4. Recover the singular solutions and apply any initial condition. Check each value where $h(y) = 0$ (the next section explains why), then, if an initial value $y(x_0) = y_0$ is given, substitute it to pin down $C$.

The geometry underneath the algebra is worth holding onto. The equation $\frac{dy}{dx} = g(x),h(y)$ assigns a slope to every point $(x, y)$ in the plane. Drawing a short line segment of that slope at many points produces a direction field, and a solution curve is simply a curve that stays tangent to the field everywhere.

Separation of variables is the algebra; the direction field is the same information drawn as a picture. Every solution you compute is one of the curves threading through that field.

Where Do Singular Solutions Come From?

Step 1 divided by $h(y)$. That move is only legal when $h(y) \neq 0$, and dividing quietly throws away any value of $y$ that makes $h(y) = 0$. Those values are not junk: each one is a genuine constant solution, called a singular or equilibrium solution, because a horizontal line $y = c$ has $\frac{dy}{dx} = 0$, which matches $g(x),h(c) = 0$ exactly when $h(c) = 0$.

The rule is short. For every root $c$ of $h(y) = 0$, the constant function $y = c$ solves the original equation. After you find the general solution by integrating, always go back and list these separately.

Sometimes the singular solution reappears as a special case of the general formula (a particular value of $C$), and sometimes it is a truly separate curve the general formula can never produce. The worked examples below show one of each, so the difference is concrete rather than abstract. This is the one step competitors skip, and it is the most common way to lose marks on an otherwise correct solution.

What Are Some Worked Separable Differential Equations Examples?

Each example is fully stepped, and every solution is verified by substituting it back into the original equation.

Example 1: Solve $\frac{dy}{dx} = xy$, then find the curve through $(0, 3)$.

Here $g(x) = x$ and $h(y) = y$. Separate the variables (assuming $y \neq 0$):

$$\frac{dy}{y} = x,dx$$

Integrate both sides:

$$\int \frac{dy}{y} = \int x,dx \quad\Longrightarrow\quad \ln\lvert y\rvert = \frac{x^2}{2} + C$$

Solve for $y$ by exponentiating. Writing $A = \pm e^{C}$ absorbs the constant and its sign:

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

Singular solution: $h(y) = y = 0$ gives $y = 0$, and here it is the special case $A = 0$, so the single family $y = A,e^{x^2/2}$ (any real $A$) captures every solution.

Now apply the initial condition $y(0) = 3$: since $e^{0} = 1$, we get $3 = A$, so $y = 3,e^{x^2/2}$.

Check by differentiating back: $\frac{dy}{dx} = 3 \cdot x,e^{x^2/2} = x\big(3,e^{x^2/2}\big) = xy$, and $y(0) = 3$. Both the equation and the condition hold.

Final answer: $y = 3,e^{x^2/2}$.

Example 2: Solve $\frac{dy}{dx} = y^2$, then find the curve through $(0, 1)$.

Now $g(x) = 1$ and $h(y) = y^2$. Separate (assuming $y \neq 0$):

$$\frac{dy}{y^2} = dx$$

Integrate both sides, using $\int y^{-2},dy = -y^{-1}$:

$$-\frac{1}{y} = x + C \quad\Longrightarrow\quad y = \frac{-1}{x + C}$$

Singular solution: $h(y) = y^2 = 0$ gives $y = 0$. Substituting $y = 0$ into the original equation gives $\frac{dy}{dx} = 0 = 0^2$, so it is a valid solution, and no finite $C$ makes $\frac{-1}{x+C}$ equal to zero. This time $y = 0$ is a true singular solution, separate from the general family.

Apply $y(0) = 1$: $1 = \frac{-1}{0 + C}$ gives $C = -1$, so $y = \frac{-1}{x - 1} = \frac{1}{1 - x}$.

Check: $\frac{dy}{dx} = \frac{d}{dx}(1 - x)^{-1} = (1 - x)^{-2}$, and $y^2 = \left(\frac{1}{1-x}\right)^2 = (1 - x)^{-2}$. They agree, and $y(0) = 1$. Note the solution blows up at $x = 1$, so its interval of validity is $x < 1$, a reminder that a solution need not exist for every $x$.

Final answer: $y = \dfrac{1}{1 - x}$ (valid for $x < 1$), with the singular solution $y = 0$.

Example 3: Solve $\frac{dy}{dx} = -\frac{x}{y}$ with $y(0) = 3$ (an implicit answer made explicit).

Here the right side factors as $g(x),h(y) = (-x)\cdot\frac{1}{y}$, so the equation is separable. Move each variable to its own side by multiplying across:

$$y,dy = -x,dx$$

Integrate both sides:

$$\frac{y^2}{2} = -\frac{x^2}{2} + C \quad\Longrightarrow\quad x^2 + y^2 = 2C$$

Writing $R^2 = 2C$, the general solution is the implicit family of circles $x^2 + y^2 = R^2$. Apply $y(0) = 3$: $0 + 9 = R^2$, so $R^2 = 9$. Solving for the branch through $(0, 3)$ gives the explicit form:

$$y = \sqrt{9 - x^2}$$

Check by implicit differentiation of $x^2 + y^2 = 9$: $2x + 2y\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{x}{y}$, the original equation, and $y(0) = \sqrt{9} = 3$. Geometrically the solution curves are circles centred at the origin, always crossing the radial direction at a right angle, which is exactly what a slope of $-\frac{x}{y}$ encodes.

Final answer: $x^2 + y^2 = 9$, or explicitly $y = \sqrt{9 - x^2}$ on $-3 < x < 3$.

How Does The Logistic Equation Use Separation Of Variables?

The logistic equation models population growth that levels off at a carrying capacity $K$:

$$\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$$

The right side is $g(t),h(P)$ with $g(t) = 1$ and $h(P) = rP\left(1 - \frac{P}{K}\right)$, so the equation is separable. Its two singular solutions come straight from $h(P) = 0$: the roots are $P = 0$ (no population) and $P = K$ (population already at capacity), and both are constant equilibrium solutions.

Separating and integrating the left side needs partial fractions, because $\frac{1}{P\left(1 - P/K\right)} = \frac{1}{P} + \frac{1/K}{1 - P/K}$. Carrying the integration through and solving for $P$ gives the S-shaped logistic curve:

$$P(t) = \frac{K}{1 + A,e^{-rt}}, \qquad A = \frac{K - P_0}{P_0}$$

As $t \to \infty$ the exponential dies and $P(t) \to K$: every non-zero starting population rises or falls toward the carrying capacity, one of the equilibria that separation flagged at the very first step. The same machinery, in its simplest form $\frac{dP}{dt} = rP$, produces plain exponential growth and decay.

When Is A First-Order Equation Separable? A Reference Table

Table: Common first-order equations, whether they separate, and why.

When an equation fails the separability test, other first-order methods take over, such as the integrating-factor method for linear differential equations, the substitution for homogeneous differential equations, or the test for exact differential equations.

Why Does Separation Of Variables Work?

Splitting $\frac{dy}{dx}$ as if it were a fraction looks like a trick, but it rests on the chain rule and the fundamental link between differentiation and integration.

• It is the chain rule read backwards. Start from $\frac{1}{h(y)}\frac{dy}{dx} = g(x)$. If $H$ is an antiderivative of $\frac{1}{h(y)}$, the left side is exactly $\frac{d}{dx},H\big(y(x)\big)$ by the chain rule. Integrating both sides with respect to $x$ gives $H(y) = \int g(x),dx + C$, which is the same relation the "separate and integrate" shortcut produces. The differentials are a bookkeeping notation for a legitimate chain-rule step.

• Each side has its own variable, so each integrates alone. Once every $y$ sits with $dy$ and every $x$ with $dx$, the two integrals are independent one-variable problems. That independence is the whole payoff of the product form $g(x),h(y)$.

• The constant carries the family. A first-order equation has a one-parameter family of solutions, and the single $+C$ is that parameter. An initial condition selects one curve from the family, which is why $C$ is found last.

Seen this way, separability is not a lucky accident of certain equations. It is the precise condition under which the chain rule can be undone one variable at a time.

Who Discovered Separable Differential Equations?

Separation of variables is almost as old as calculus itself, arriving within a few years of the notation that made it writable.

Two figures anchor the story with dates:

• Gottfried Wilhelm Leibniz (1646–1716, Germany) introduced the differential notation and the separation technique that still bears no better name than "separation of variables."

• Johann Bernoulli (1667–1748, Switzerland) turned the method into a working tool for applied problems and taught it to the next generation, including Euler.

Later, rigour caught up: the existence-and-uniqueness theorems of the nineteenth century explained when the one-parameter family and its singular solutions form the complete answer.

Where Are Separable Differential Equations Used In The Real World?

Any process whose rate of change depends on the current amount tends to be separable, which is why the method reaches across the sciences.

• Radioactive decay and cooling: $\frac{dy}{dx} = ky$ models decay, and Newton's law of cooling, $\frac{dT}{dt} = k(T - T_{\text{room}})$, both separate and integrate to exponential curves.

• Population biology: unchecked growth is $\frac{dP}{dt} = rP$; growth with a ceiling is the logistic equation, and both are solved by separation.

• Draining tanks (Torricelli's law): the height of water in a leaking tank obeys $\frac{dh}{dt} = -k\sqrt{h}$, a separable equation that predicts the emptying time.

• Mixing problems: the concentration of salt in a stirred tank with inflow and outflow leads to a separable first-order equation in chemical engineering.

• Simple circuits: the charge on a capacitor in an RC circuit, $\frac{dq}{dt} = \frac{1}{R}\left(V - \frac{q}{C}\right)$, separates to give the familiar charging curve.

One method, "separate and integrate," quietly powers models of decay, populations, fluids, and electronics, which is why it is the first solution technique every differential-equations course teaches.

What Are The Most Common Mistakes With Separable Differential Equations?

These four errors account for most lost marks, and each matches a question real students ask on r/calculus, r/learnmath, and course common-error handouts.

Losing the singular solution when you divide by $h(y)$.

Where it slips in:

Solving $\frac{dy}{dx} = y^2$, a student divides by $y^2$, integrates to $y = \frac{-1}{x + C}$, and stops, never noticing that $y = 0$ was discarded at the division step.

Don't do this:

Do not treat the general family as the whole answer. Dividing by $h(y)$ silently removes every $y$ with $h(y) = 0$.

The correct way:

Before or after integrating, solve $h(y) = 0$ and test each root in the original equation. For $\frac{dy}{dx} = y^2$, the root $y = 0$ is a valid constant solution that must be stated alongside the family.

Adding the constant of integration too late, or to both sides.

Where it slips in:

A student integrates, solves all the way to an explicit $y$, and only then appends $+C$, or writes a separate constant on each side and never combines them.

Don't do this:

Do not attach $C$ after solving for $y$, and do not carry two constants. Adding $+C$ at the wrong stage changes the family of curves entirely.

The correct way:

Add a single $+C$ the moment you integrate, while the equation is still $\int \frac{dy}{h(y)} = \int g(x),dx + C$, then carry it through the algebra. Two constants, one from each side, always merge into one.

Mishandling the logarithm-to-exponential step.

Where it slips in:

From $\ln\lvert y\rvert = \frac{x^2}{2} + C$, a student writes $y = e^{x^2/2} + e^{C}$, splitting the exponential of a sum into a sum of exponentials, or drops the absolute value and loses the negative branch.

Don't do this:

Do not turn $e^{a + b}$ into $e^{a} + e^{b}$, and do not ignore the sign hidden by $\lvert y\rvert$.

The correct way:

Use $e^{a + b} = e^{a}e^{b}$, so $e^{,x^2/2 + C} = e^{C}e^{x^2/2}$, and fold the constant sign into one symbol: $y = A,e^{x^2/2}$ with $A = \pm e^{C}$ any real constant.

Trying to separate an equation that is not separable.

Where it slips in:

Faced with $\frac{dy}{dx} = x + y$, a student forces a split such as $\frac{dy}{y} = \frac{dx}{x}$, which does not follow from the original equation at all.

Don't do this:

Do not separate before checking that the right side factors as $g(x),h(y)$. A sum like $x + y$ never factors that way.

The correct way:

Test the form first. If it will not factor into an $x$-only part times a $y$-only part, reach for a linear method (integrating factor), a homogeneous substitution, or an exactness test instead.

Practice Problems On Separable Differential Equations

Work each one, then check against the answer. Every answer has been verified by substitution.

1. Solve $\frac{dy}{dx} = \frac{y}{x}$ for $x > 0$.
   (Answer: $\frac{dy}{y} = \frac{dx}{x}$ gives $\ln\lvert y\rvert = \ln\lvert x\rvert + C$, so $y = Ax$; singular solution $y = 0$ is the case $A = 0$.)

2. Solve $\frac{dy}{dx} = 2xy$ with $y(0) = 5$.
   (Answer: $\ln\lvert y\rvert = x^2 + C$, so $y = 5e^{x^2}$.)

3. Solve $\frac{dy}{dx} = e^{x - y}$.
   (Answer: $e^{y},dy = e^{x},dx$, so $e^{y} = e^{x} + C$, giving $y = \ln\left(e^{x} + C\right)$.)

4. Solve $\frac{dy}{dx} = \frac{3x^2}{y^2}$.
   (Answer: $y^2,dy = 3x^2,dx$, so $\frac{y^3}{3} = x^3 + C$, giving $y = \left(3x^3 + C_1\right)^{1/3}$.)

5. Solve $\frac{dy}{dx} = y - 1$ with $y(0) = 3$.
   (Answer: $\ln\lvert y - 1\rvert = x + C$, so $y = 1 + 2e^{x}$; singular solution $y = 1$.)

6. Solve $\frac{dy}{dx} = \frac{\cos x}{y}$ with $y(0) = 2$.
   (Answer: $y,dy = \cos x,dx$, so $\frac{y^2}{2} = \sin x + 2$, giving $y = \sqrt{2\sin x + 4}$.)

Where Should You Go Next After Separable Differential Equations?

Separation of variables is the first door into differential equations, and several natural next steps open from here.

1. Ordinary differential equations. See where separable equations sit in the wider map of first- and higher-order equations.

2. Linear differential equations. Learn the integrating-factor method for the many equations that are not separable.

3. Initial value problems. Go deeper on using a condition $y(x_0) = y_0$ to select one curve from the solution family.

If your child is meeting separable differential equations for the first time, a live Bhanzu trainer teaches the method from the direction-field picture up, so separation feels like following a slope rather than shuffling symbols, in the Bhanzu math classes.