Homogeneous First-Order Reduction: Substitute y equals v times x

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: after recognizing homogeneous first-order form, set $y=vx$, treat $v$ as a function of $x$, and rewrite the differential equation in terms of $v$ and $x$.

The invariant: the substitution re-expresses the same solution branch on a nonzero-$x$ interval; it changes coordinates from $y$ to the ratio variable $v=y/x$.

Pattern:

$$y^{\prime}=F(y/x),\quad y=vx \quad \longrightarrow \quad xv^{\prime}=F(v)-v$$

Dividing by $x$ then gives $v^{\prime}=\frac{F(v)-v}{x}$, which is why this reduction usually prepares the equation for separation.

Legal route	Illegal route
$y^{\prime}=1+\frac{y}{x},\ y=vx \Longrightarrow v+xv^{\prime}=1+v$	$y^{\prime}=x+\left(\frac{y}{x}\right)^2,\ y=vx \not\Longrightarrow xv^{\prime}=F(v)-v$; substitution gives $v+xv^{\prime}=x+v^2$, so direct $x$ dependence remains.

The illegal column is not illegal algebra; it is illegal route selection. The substitution can still be performed, but it does not produce the canonical homogeneous-reduction form because direct $x$ dependence remains.

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Conditions of Applicability

Condition: $homogeneous first-order form y^{\prime}=F(y/x)$; $x\neq 0 on working interval$

Before applying, check: first confirm Homogeneous First-Order Equation Form: the entire right-hand side must depend on $y/x$ alone.

If the condition is violated: the substitution may leave direct $x$ or $y$ dependence that does not collapse to $xv^{\prime}=F(v)-v$, so the homogeneous-reduction route is not licensed.

• Work on an interval that does not cross $x=0$, because $v=y/x$ must be defined.
• After substituting $y=vx$, use the product rule $y^{\prime}=v+xv^{\prime}$.
• If the right-hand side contains direct $x$ or $y$ dependence after the ratio check, choose another first-order route instead of forcing this reduction.

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

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Common Failure Modes

Failure mode: see a term $y/x$ and apply $y=vx$ as a homogeneous reduction without checking the whole right-hand side -> the result still contains direct $x$ or $y$ dependence.

Debug: replace every $y/x$ by $v$ mentally; if direct $x$ or $y$ dependence remains in the slope rule, the homogeneous reduction is not the right route.

Failure mode: set $y=vx$ but differentiate as if $v$ were constant -> the required $xv^{\prime}$ term disappears.

Debug: say out loud that $v=v(x)$ before differentiating, then use $y^{\prime}=v+xv^{\prime}$.

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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 $y=vx$ turn the ratio $y/x$ into the new variable $v$?
• Why must the derivative become $y^{\prime}=v+xv^{\prime}$ instead of only $v$?

For the Principle

• What quick check tells you whether the right-hand side is truly a function of $y/x$ alone?
• What breaks in the reduction if the working interval crosses $x=0$?

Between Principles

• How does this reduction follow after Homogeneous First-Order Equation Form and differ from Separation of Variables?

Generate an Example

• Create one eligible equation of the form $y^{\prime}=F(y/x)$ and one near miss that contains $y/x$ plus direct $x$ dependence.

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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: _____Set y equal to v times x in a homogeneous first-order equation, then rewrite the equation as a reduced differential equation for v as a function of x.

Write the canonical pattern: _____$y^{\prime}=F(y/x),\ y=vx \Rightarrow xv^{\prime}=F(v)-v$

State the canonical condition: _____$homogeneous first-order form y^{\prime}=F(y/x); x\neq 0 on working interval$

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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 $y^{\prime}=1+\frac{y}{x}$ on an interval where $x>0$, reach the reduced equation for $v$ using $y=vx$.

Step	Expression	Operation
0	$y^{\prime}=1+\frac{y}{x}$	Start from homogeneous first-order form on a nonzero-$x$ interval.
1	$y=vx,\quad \frac{y}{x}=v$	Introduce the ratio variable.
2	$y^{\prime}=v+xv^{\prime}$	Differentiate $y=vx$ with $v=v(x)$.
3	$v+xv^{\prime}=1+v$	Substitute into the original equation.
4	$xv^{\prime}=1$	Subtract $v$ to isolate the reduced differential equation.

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Drills

Goal Micro-Chain

Reduce the equation using $y=vx$. Assume the equation has homogeneous first-order form and $x>0$.

$$y^{\prime}=2+\frac{y}{x}$$

Reveal

Step	Expression	Move
0	$y^{\prime}=2+\frac{y}{x}$	Check that the slope is $F(y/x)$ with $F(v)=2+v$.
1	$y=vx,\quad y^{\prime}=v+xv^{\prime}$	Substitute and differentiate.
2	$v+xv^{\prime}=2+v$	Replace $y/x$ by $v$.
3	$xv^{\prime}=2$	Subtract $v$.

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Reduce the equation using $y=vx$. Assume $x<0$ on the working interval.

$$y^{\prime}=\left(\frac{y}{x}\right)^2$$

Reveal

Step	Expression	Move
0	$y^{\prime}=\left(\frac{y}{x}\right)^2$	The right-hand side is $F(y/x)$ with $F(v)=v^2$.
1	$y=vx,\quad y^{\prime}=v+xv^{\prime}$	Use the product rule for $v(x)x$.
2	$v+xv^{\prime}=v^2$	Replace $y/x$ by $v$.
3	$xv^{\prime}=v^2-v$	Subtract $v$.

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Reject or complete the route choice. Assume $x>0$.

$$y^{\prime}=x+\left(\frac{y}{x}\right)^2$$

Reveal

Reject the homogeneous reduction route. The right-hand side contains direct $x$ dependence, so it is not a function of $y/x$ alone.

If you substitute anyway, you get

$$v+xv^{\prime}=x+v^2,$$

which is not the canonical reduced form $xv^{\prime}=F(v)-v$.

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Reduce the equation using $y=vx$. Assume the equation has homogeneous first-order form and $x>0$.

$$y^{\prime}=\frac{y}{x}-3$$

Reveal

Step	Expression	Move
0	$y^{\prime}=\frac{y}{x}-3$	Here $F(v)=v-3$.
1	$v+xv^{\prime}=v-3$	Substitute $y=vx$ and $y^{\prime}=v+xv^{\prime}$.
2	$xv^{\prime}=-3$	Subtract $v$.

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Forward Step

Apply the substitution step once. Assume $x\neq 0$ on the interval.

$$y=vx,\quad y^{\prime}=4+\frac{y}{x}$$

Reveal

Because $v=v(x)$ and $y/x=v$:

$$v+xv^{\prime}=4+v$$

Then the reduced equation is $xv^{\prime}=4$.

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Apply the substitution step once. Assume $x\neq 0$ on the interval.

$$y=vx,\quad y^{\prime}=1+\left(\frac{y}{x}\right)^2$$

Reveal

$$v+xv^{\prime}=1+v^2$$

so

$$xv^{\prime}=1+v^2-v.$$

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Which equation is eligible for homogeneous first-order reduction on an interval where $x>0$?

A. $y^{\prime}=\sin(y/x)+2$
B. $y^{\prime}=\sin(y/x)+x$

Reveal

A is eligible because the whole right-hand side is a function of $y/x$ alone: $F(v)=\sin v+2$.

B is not eligible because the direct $x$ term remains after $y/x$ becomes $v$.

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A student writes the derivative step below. What is wrong?

$$y=vx \quad \Longrightarrow \quad y^{\prime}=v$$

Reveal

The step treats $v$ as a constant. In this reduction, $v$ is a function of $x$, so the product rule gives

$$y^{\prime}=v+xv^{\prime}.$$

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Transition Identification

What move happened in this transition? Assume $x>0$.

$$y^{\prime}=5+\frac{y}{x} \quad \longrightarrow \quad v+xv^{\prime}=5+v$$

Reveal

The homogeneous reduction substitution was applied: $y=vx$, so $y/x=v$ and $y^{\prime}=v+xv^{\prime}$.

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What is missing from this worked chain?

$$y^{\prime}=1+\frac{y}{x} \quad \longrightarrow \quad v+xv^{\prime}=1+v$$

Reveal

The chain should state the condition: the equation has homogeneous first-order form, and the working interval must have $x\neq 0$ so $v=y/x$ is defined.

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Starting from $y^{\prime}=1+\left(\frac{y}{x}\right)^2-\frac{y}{x}$ on an interval where $x>0$, reach the reduced equation for $v$ using Homogeneous First-Order Reduction.

Full solution

Step	Expression	Move
0	$y^{\prime}=1+\left(\frac{y}{x}\right)^2-\frac{y}{x}$	Confirm homogeneous first-order form on a nonzero-$x$ interval.
1	$y=vx,\quad \frac{y}{x}=v$	Introduce the ratio variable.
2	$y^{\prime}=v+xv^{\prime}$	Differentiate $y=vx$ with $v=v(x)$.
3	$v+xv^{\prime}=1+v^2-v$	Substitute into the equation.
4	$xv^{\prime}=1+v^2-2v$	Subtract $v$ from both sides.

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Related Guides

• Differential Equations Subdomain - See where homogeneous reduction sits in the first-order ODE sequence.
• Homogeneous First-Order Equation Form - Check the ratio-dependent slope form before making this substitution.
• First-Order Explicit Differential Equation Form - Start from the broader explicit first-order pattern before choosing a solver route.
• Separation of Variables - Compare this reduction route with the later separated-variable rewrite used for different structures.
• Principle Structures - Track names, conditions, formulas, and neighboring moves as separate recall targets.

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FAQ

What is Homogeneous First-Order Reduction?

Homogeneous First-Order Reduction is the substitution $y=vx$ applied to a first-order ODE whose slope depends only on $y/x$. It rewrites $y^{\prime}=F(y/x)$ as $xv^{\prime}=F(v)-v$, an equation for the ratio variable $v(x)$.

When is the substitution y equals v x valid?

It is valid for this reduction when the equation has homogeneous first-order form and the working interval avoids $x=0$. The nonzero-$x$ condition makes $v=y/x$ meaningful on the whole interval being solved.

Why does y prime become v plus x v prime?

Because $v$ is not a constant. The substitution means $y=v(x)x$, so the product rule gives $y^{\prime}=v+xv^{\prime}$.

Is homogeneous reduction the same as separation of variables?

No. Homogeneous reduction changes variables from $y$ to $v=y/x$. The reduced equation may then become separable, but that is a later route decision, not the same move.

What is the fastest way to reject a near miss?

Replace $y/x$ by $v$ in the slope rule. If direct $x$ or $y$ dependence remains, the equation is not in homogeneous first-order form and the homogeneous reduction route is not licensed.

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How This Fits in Unisium

Within the differential equations subdomain, Unisium treats homogeneous reduction as a route-selection move: recognize the ratio-form equation, check the nonzero-$x$ interval, then execute $y=vx$ without dropping the product-rule term. The Unisium Study System pairs that condition check with retrieval practice, self-explanation, and compact problem-solving chains so the substitution becomes fluent without hiding the legality check.

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