Selecting a Solution Method for First-Order Differential Equations

date2026-07-16document_iddoc_88992b92abd9529a5319c2142a42cec2description一階微分方程式が複数の型に重なることを前提に、定義域と定数解を保全して最も軽い解法を選ぶ診断手順を整理する。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 / 一階線型微分方程式と積分因子 / 完全微分方程式 / 同次形一階微分方程式と置換 / Bernoulli 方程式 / 積分法の基本type講義content_typelecturestatusactive

mathdifferential-equationsfirst-orderlecture

1Introduction

This lecture explains how to identify the structure of a first-order differential equation and select a solution method, beginning with forms that require the least transformation. Because one equation may belong to several classes, the diagnostic tree is a priority order rather than an exclusive classification.

2Rationale for the Diagnostic Order

The order is determined by the amount of algebraic transformation required to recognize each structure. Separable and autonomous equations can be identified directly from factorization and dependence of the right-hand side. A first-order linear equation is identified by checking whether $y$ and $y^{'}$ occur linearly, with coefficients depending only on $x$ . Exact differential equations, Bernoulli equations, and homogeneous first-order equations are considered afterward because they require conversion to a differential form or an appropriate substitution.

This order prevents a simple structure from being obscured by an unnecessarily complicated substitution. Diagnosis is a preliminary step before computation; it is not itself a solution method.

3Diagnostic Tree

Determine the domain of the right-hand side and identify every factor that may vanish before dividing by it.
If $y^{'} = F (y)$ , treat the equation as autonomous and first preserve every equilibrium solution $y = c$ satisfying $F (c) = 0$ . The non-equilibrium branches are separable.
If the equation can be written as $y^{'} = f (x) g (y)$ , use separation of variables as the first candidate, while retaining constant solutions for which $g (y) = 0$ .
If the equation can be written as $y^{'} + P (x) y = Q (x)$ , use an integrating factor. If it is also separable, choose the method with the shorter computation.
If the equation has the form $M (x, y) d x + N (x, y) d y = 0$ , check the exactness condition $M_{y} = N_{x}$ together with the domain conditions.
If the equation has the form $y^{'} + P (x) y = Q (x) y^{n}$ , consider the Bernoulli substitution $u = y^{1 - n}$ .
If the equation has the form $y^{'} = F (y / x)$ , consider the substitution $v = y / x$ on an interval where $x \neq 0$ .

4Classification Table

Type	Diagnostic feature	Typical misclassification
Separable	Can be factored into an $x$ -dependent factor and a $y$ -dependent factor	Treating $x + y$ as though it were a product
First-order linear	$y$ and $y^{'}$ occur linearly, with coefficients depending only on $x$	Classifying an equation containing $y y^{'}$ or $y^{2}$ as linear
Exact differential equation	Has the differential form $M d x + N d y$	Omitting the domain conditions
Bernoulli	Contains $y^{n}$ and can be linearized by substitution	Excluding it merely because it is nonlinear
Homogeneous first-order	Depends on $y / x$	Confusing it with a homogeneous linear equation

5Checks for Each Branch

Branch	Structural feature to verify	Common error
Separable	$y^{'} = f (x) g (y)$ or $A (y) y^{'} = B (x)$	Mistaking $x + y$ for $f (x) g (y)$
First-order linear	$y^{'}$ and $y$ occur to the first power, with coefficients depending only on $x$	Classifying an equation containing $y y^{'}$ or $y^{2}$ as linear
Exact differential equation	$M d x + N d y = 0$ with $M_{y} = N_{x}$	Ignoring holes in the domain or a sufficient global condition such as simple connectivity
Bernoulli	$y^{'} + P (x) y = Q (x) y^{n}$	Treating the degenerate cases $n = 0, 1$ as ordinary Bernoulli cases
Homogeneous first-order	$y^{'} = F (y / x)$	Confusing it with the homogeneous linear equation $L [y] = 0$

6Examples

The equation

y^{'} = x y

is separable and is also first-order linear in the form $y^{'} - x y = 0$ . First retain the constant solution $y = 0$ . On the branch $y \neq 0$ , separation of variables is preferable because it requires fewer steps.

The equation

y^{'} + x y = 1

is not separable; it is first-order linear with $P (x) = x$ and $Q (x) = 1$ . An integrating factor is therefore appropriate.

7Example in Which the Classification Changes after Substitution

Consider

y^{'} = \frac{y}{x} + {(\frac{y}{x})}^{2}, x \neq 0,

on an interval contained in either $x > 0$ or $x < 0$ . Set $v = y / x$ , so that $y = v x$ and $y^{'} = v + x v^{'}$ . Substitution gives

v + x v^{'} = v + v^{2},

which becomes the separable equation $x v^{'} = v^{2}$ . Before division by $v^{2}$ , also retain the solution $v = 0$ , corresponding to $y = 0$ .

In this example, dependence on $x$ and $y$ only through the ratio $y / x$ justifies the homogeneous substitution. After substitution, one must again check for branches that division would discard. This calculation does not justify a solution that crosses $x = 0$ .

8When No Elementary Branch Applies

An equation such as

y^{'} = \sin (x y)

does not directly belong to any of the elementary classes above. The right-hand side $f (x, y) = \sin (x y)$ and its partial derivative with respect to $y$ ,

f_{y} (x, y) = x \cos (x y),

are continuous. A subsequent lecture on existence and uniqueness will use continuity of $f$ to establish local existence and local Lipschitz continuity in $y$ to establish uniqueness. In this situation, it is appropriate to proceed from the search for an elementary solution to direction fields, numerical methods, and qualitative analysis introduced in subsequent lectures.

9Scope

This diagnostic tree is an entry point to elementary methods. Failure to match any branch does not imply that the equation has no solution. It may instead indicate that numerical methods or qualitative analysis are required.

10Exercises

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

11Next Reference

data/lecture/math/differential-equations/lipschitz-condition-and-continuity.lecture.n.md