Homogeneous First-Order Differential Equations and Substitution

date2026-07-16document_iddoc_5f819de6f12e89e191c3b5195bc234f1description一階微分方程式の同次形を、比 y/x を未知関数にする置換で変数分離形へ帰着する定石として証明付きで整理する。prerequisites一階微分方程式の分類と最初の判定 / 変数分離形と自律方程式type講義content_typelecturestatusactive

mathdifferential-equationsfirst-ordersubstitutionlecture

1Introduction

This lecture explains how the substitution

v = \frac{y}{x}

reduces a homogeneous first-order differential equation to a separable equation. Here, homogeneous means dependence on the ratio $y / x$ ; it is distinct from the term homogeneous used for a second-order linear equation with zero right-hand side.

2Standard Form

A homogeneous first-order equation can be written as

y^{'} = F (\frac{y}{x}) .

Because $y / x$ is undefined at $x = 0$ , solutions are considered on intervals contained in either $x > 0$ or $x < 0$ .

3Introducing the Substitution $v = y / x$

Since the right-hand side depends only on $y / x$ , define

v (x) = \frac{y (x)}{x} .

Then $y = x v$ , and the product rule gives

y^{'} = v + x v^{'} .

Substitution converts the original equation into an equation involving $v$ and $x$ .

4Proposition: Reduction to a Separable Equation

Consider

y^{'} = F (\frac{y}{x})

on an interval where $x \neq 0$ . With $v = y / x$ , the function $v$ satisfies

x v^{'} = F (v) - v .

On a range where $F (v) - v \neq 0$ , this becomes the separable equation

\frac{d v}{F (v) - v} = \frac{d x}{x} .

4.1Proof

From $y = x v$ and the product rule,

y^{'} = v + x v^{'} .

Substitution gives

v + x v^{'} = F (v),

and hence $x v^{'} = F (v) - v$ . Since $x \neq 0$ , division by $x$ is valid; where $F (v) - v \neq 0$ , the variables can also be separated.

Before division by $F (v) - v$ , check its zeros. If $F (a) - a = 0$ , then $v = a$ and therefore $y = a x$ is a straight-line solution.

5Example: $y^{'} = 1 + y / x$

Set $v = y / x$ and $y = x v$ . Then

v + x v^{'} = 1 + v,

so $x v^{'} = 1$ and

v^{'} = \frac{1}{x} .

Integration yields

v = \log | x | + C .

Returning to $y = x v$ gives

y = x (\log | x | + C) .

6An Equation Outside the Scope

The equation

y^{'} = x + y

cannot be expressed as a function of $y / x$ alone, so the substitution $v = y / x$ does not apply. Rewriting it as

y^{'} - y = x

shows that it is a first-order linear equation.

data/lecture/math/differential-equations/first-order-linear-odes-and-integrating-factors.lecture.n.md

7Summary

Determine whether the equation can be written as $y^{'} = F (y / x)$ .
Work on an interval that excludes $x = 0$ .
Set $v = y / x$ , equivalently $y = x v$ .
Substitute $y^{'} = v + x v^{'}$ .
Check constant roots of $F (v) - v$ before division, then reduce the remaining equation to separable form.

The next lecture treats Bernoulli equations, where substitution of $y^{1 - n}$ reduces a nonlinear equation to a first-order linear equation.

data/lecture/math/differential-equations/bernoulli-equations.lecture.n.md

8Exercises

data/exercise/math/differential-equations/classifying-and-solving-first-order-odes.exercise.n.md