Separation of Variables: The Safe Rewrite Step

On this page: The Principle | Conditions | Failure Modes | EE Questions | Retrieval Practice | Practice Ground | Solve a Problem | Related Guides | FAQ

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The Principle

The move: rewrite a separable equation into separated differential form so every $y$-dependent factor sits with $dy$ and every $x$-dependent factor sits with $dx$.

What stays true: on any interval where $h(y(x))\neq 0$, the separated equation describes the same non-equilibrium solution branch as the original equation.

Pattern:

$$y^{\prime}=g(x)h(y) \quad \longrightarrow \quad \frac{1}{h(y)}\,dy=g(x)\,dx$$

Here $y^{\prime}$ means $\frac{dy}{dx}$, so the rewrite moves from derivative form to separated differential form, ready for integration.

Complete	Incomplete
$y^{\prime}=x(y-3)$ gives the equilibrium branch $y=3$, and on $y\neq 3$: $\frac{1}{y-3}\,dy=x\,dx$	$y^{\prime}=x(y-3) \longrightarrow \frac{1}{y-3}\,dy=x\,dx$ presented as the whole solution set, without checking $y=3$

The complete column says which branch is being solved. The incomplete column treats division by $h(y)$ as harmless everywhere, which can erase an equilibrium solution.

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Conditions of Applicability

Condition: $h(y)\neq 0$; separable product form; equilibrium branch handled separately

Before applying, check: first confirm separable equation product form, then find zeros of $h(y)$ before dividing by it.

If the condition is violated: dividing by $h(y)$ can delete constant solutions where $h(y)=0$, or apply the solver route to an equation that is not separable.

• The equation must already be in first-order explicit product form $y^{\prime}=g(x)h(y)$.
• On the branch you rewrite, $h(y(x))$ must stay nonzero.
• Any constant solution $y=c$ with $h(c)=0$ belongs to a separate equilibrium branch, connected to Scalar Equilibrium Solution Condition.

In practice, the separated equation describes a solution branch on an interval where the chosen solution does not cross a zero of $h(y)$.

Want the complete framework behind this guide? Read Masterful Learning.

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Common Failure Modes

Failure mode: divide by $h(y)$ without checking whether $h(y)=0$ has constant solutions -> an equilibrium branch disappears from the answer.

Debug: set $h(y)=0$ before separating variables, record each equilibrium solution, then solve the non-equilibrium branch.

Failure mode: treat a right-hand side such as $x+y$ as separable because it contains both variables -> the variables cannot be separated by this rewrite.

Debug: ask whether the right-hand side is an $x$-only factor times a $y$-only factor, not a sum or mixed expression.

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• Why does the reciprocal factor $1/h(y)$ belong on the $dy$ side after rewriting $y^{\prime}=\frac{dy}{dx}$?
• What does the phrase “non-equilibrium branch” mean when $h(y)=0$ has roots?

For the Principle

• What exact check must happen before dividing by $h(y)$?
• How would the rewrite fail if $y^{\prime}=x+y$ were treated as if it had product form?

Between Principles

• How does this rewrite follow after recognizing Separable Equation Product Form?

Generate an Example

• Create one separable equation with an equilibrium branch and one near-miss equation that cannot use the separation rewrite.

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the move in one sentence: _____Rewrite a separable first-order equation into separated differential form by dividing by the nonzero y-factor and pairing it with dy.

Write the canonical pattern: _____$y^{\prime}=g(x)h(y) \Rightarrow \frac{1}{h(y)}\,dy=g(x)\,dx$

State the canonical condition: _____$h(y)\neq 0;\, \text{separable product form};\, \text{equilibrium branch handled separately}$

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Practice Ground

Use these exercises to build move-selection fluency. (See Self-Explanation for how to learn from worked examples.)

Procedure Walkthrough

Starting from $y^{\prime}=x(y-3)$, reach separated differential form while keeping the equilibrium branch visible.

Step	Expression	Operation
0	$y^{\prime}=x(y-3)$	Start from separable product form with $g(x)=x$ and $h(y)=y-3$.
1	$y=3$	Check the equilibrium branch from $h(y)=0$.
2	$y\neq 3,\quad y^{\prime}=x(y-3)$	Restrict to a non-equilibrium branch before division.
3	$y\neq 3,\quad \frac{1}{y-3}\,dy=x\,dx$	Divide by the nonzero $y$-factor and separate variables.

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Drills

Forward Step

Apply the separation rewrite once. State any equilibrium branch first.

$$y^{\prime}=2x(y+1)$$

Reveal

Equilibrium branch: $y=-1$.

On $y\neq -1$:

$$\frac{1}{y+1}\,dy=2x\,dx$$

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Apply the separation rewrite once. State any equilibrium branch first.

$$y^{\prime}=e^x y^2$$

Reveal

Equilibrium branch: $y=0$.

On $y\neq 0$:

$$\frac{1}{y^2}\,dy=e^x\,dx$$

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Which equation is eligible for the separation rewrite?

A. $y^{\prime}=x(y-4)$
B. $y^{\prime}=x+y-4$

Reveal

A is eligible because the right-hand side is an $x$-only factor times a $y$-only factor. Its equilibrium branch is $y=4$, and the rewrite on $y\neq 4$ is

$$\frac{1}{y-4}\,dy=x\,dx$$

B is not eligible because $x+y-4$ is a sum, not separable product form.

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A student proposes the rewrite below. Is it valid as written?

$$y^{\prime}=x(y-2) \quad \longrightarrow \quad \frac{1}{y-2}\,dy=x\,dx$$

Reveal

It is incomplete as written. The non-equilibrium rewrite is valid only on $y\neq 2$, and the equilibrium branch $y=2$ must be handled separately.

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Goal Micro-Chain

Reach separated differential form from the start equation.

$$y^{\prime}=(x^2+1)\sin y$$

Reveal

Step	Expression	Move
0	$y^{\prime}=(x^2+1)\sin y$	Identify $g(x)=x^2+1$ and $h(y)=\sin y$.
1	$y=k\pi$ for any integer $k$	Record equilibrium branches from $\sin y=0$.
2	$\sin y\neq 0$	Work on a non-equilibrium branch.
3	$\frac{1}{\sin y}\,dy=(x^2+1)\,dx$	Divide by the nonzero $y$-factor.

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Reach separated differential form from the start equation.

$$y^{\prime}=\frac{x}{1+x^2}(1-y)$$

Reveal

Step	Expression	Move
0	$y^{\prime}=\frac{x}{1+x^2}(1-y)$	Identify $g(x)=\frac{x}{1+x^2}$ and $h(y)=1-y$.
1	$y=1$	Record the equilibrium branch.
2	$y\neq 1$	Work on a branch where $1-y$ is nonzero.
3	$\frac{1}{1-y}\,dy=\frac{x}{1+x^2}\,dx$	Separate variables.

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Reject or complete the route choice.

$$y^{\prime}=xy+y$$

Reveal

First factor the right-hand side:

$$y^{\prime}=y(x+1)$$

Now the equation has separable product form with $g(x)=x+1$ and $h(y)=y$. The equilibrium branch is $y=0$, and on $y\neq 0$:

$$\frac{1}{y}\,dy=(x+1)\,dx$$

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Reject or complete the route choice.

$$y^{\prime}=xy+1$$

Reveal

Reject this route. The right-hand side is not an $x$-only factor times a $y$-only factor, so the separation rewrite is not licensed.

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Transition Identification

Where did the separation rewrite occur?

$$y^{\prime}=3x(y^2+1) \quad \longrightarrow \quad \frac{1}{y^2+1}\,dy=3x\,dx$$

Reveal

The rewrite occurs in the single transition. Since $y^2+1$ is never zero for real $y$, no real equilibrium branch comes from the $y$-factor.

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What is missing from this worked chain?

$$y^{\prime}=(x-1)(y+5) \quad \longrightarrow \quad \frac{1}{y+5}\,dy=(x-1)\,dx$$

Reveal

The chain omitted the equilibrium branch $y=-5$ and the branch condition $y\neq -5$ before division.

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Starting from $y^{\prime}=x^2(y-6)$, reach separated differential form and preserve every branch that must be considered.

Full solution

Step	Expression	Move
0	$y^{\prime}=x^2(y-6)$	Start from separable product form with $g(x)=x^2$ and $h(y)=y-6$.
1	$y=6$	Record the equilibrium branch from $h(y)=0$.
2	$y\neq 6,\quad y^{\prime}=x^2(y-6)$	Work on a non-equilibrium branch before division.
3	$y\neq 6,\quad \frac{1}{y-6}\,dy=x^2\,dx$	Divide by the nonzero $y$-factor and separate variables.
4	$\int \frac{1}{y-6}\,dy=\int x^2\,dx$	The result is ready for integration on the non-equilibrium branch.

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Related Guides

• Differential Equations Subdomain - See where separable-variable rewriting sits in the first-order ODE sequence.
• Separable Equation Product Form - Check the structure before making this rewrite.
• Scalar Equilibrium Solution Condition - Keep constant solutions visible before dividing by a $y$-factor.
• Principle Structures - Track names, formulas, and conditions as distinct pieces of knowledge.

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FAQ

What is separation of variables in differential equations?

Separation of variables is the rewrite from $y^{\prime}=g(x)h(y)$ to $\frac{1}{h(y)}\,dy=g(x)\,dx$ on a branch where $h(y)\neq 0$. In Unisium, this specific step is treated as Separable Variable Separation Rewrite: the move that makes a separable equation ready for integration while keeping equilibrium branches visible.

When is the separation rewrite valid?

It is valid when the equation has separable product form, the branch being solved has $h(y)\neq 0$, and any equilibrium solution from $h(y)=0$ has already been handled separately.

Why do I have to check equilibrium solutions first?

Dividing by $h(y)$ is not legal on a branch where $h(y)=0$. If $h(c)=0$, then $y=c$ may be a constant solution, and that branch can vanish if you divide by $h(y)$ too early.

Is recognizing separable product form the same as separating variables?

No. Recognizing product form tells you the equation has the shape $y^{\prime}=g(x)h(y)$. Separating variables is the later rewrite that moves from that shape to $\frac{1}{h(y)}\,dy=g(x)\,dx$ after the branch check.

Does this rewrite solve the differential equation?

No. It prepares the equation for integration. After the rewrite, you still integrate both sides and handle constants, intervals, and any initial condition.

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How This Fits in Unisium

Within the differential equations subdomain, Unisium treats this as a route-selection move: first recognize separable product form, then decide whether the division by $h(y)$ is legal on the branch you are solving. The Unisium Study System pairs that condition check with retrieval practice, self-explanation, and compact problem-solving chains so the rewrite becomes fluent without hiding the equilibrium branch.

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