Classification and Initial Diagnosis of First-Order Differential Equations

1Introduction

This lecture explains how to analyze the form of a first-order differential equation and identify both an applicable solution method and the reason for selecting it.

A first-order differential equation is a differential equation whose highest derivative is $$y^{\prime }$$y'. It can generally be written as

There is no universal formula that solves this general form. The initial objective is therefore to recognize frequently occurring classes and explain why their structure permits a particular method.

2Three Initial Checks

Before calculating, check the following features of a first-order equation.

1. Is the right-hand side a function of $$x$$x alone?
2. Can the dependence on $$x$$x and $$y$$y be separated into factors?
3. Do $$y^{\prime }$$y' and $$y$$y occur linearly?

Whenever division by a factor is contemplated, also determine whether setting that factor equal to zero yields a solution. First verify the zero-factor branch by substitution; only then analyze branches on which the factor is nonzero. Distinguishing envelope-type singular solutions from nonunique solution branches is a later topic that depends on existence and uniqueness theory.

This order is efficient because simpler classes can be identified with fewer transformations. Prematurely attempting a complicated substitution may obscure an equation that is solvable by direct integration or separation of variables.

3Classification Table

The table is not intended for memorization in isolation. Its purpose is to explain the correspondence between equation structure and solution method.

4Proposition 1: $$y^{\prime }=f\left(x\right)$$y'=f(x) Reduces to Direct Integration

Suppose that

and that $$f$$f is continuous on an interval $$I$$I. If $$F$$F is an antiderivative of $$f$$f on $$I$$I, then

is a solution for every constant $$C$$C.

4.1Proof

By definition, $$F^{\prime }=f$$F'=f. If $$y=F\left(x\right)+C$$y=F(x)+C, then the derivative of $$C$$C is zero, so

Thus $$y=F\left(x\right)+C$$y=F(x)+C satisfies the equation. In this class, solving the differential equation is precisely the problem of determining antiderivatives.

5Proposition 2: $$y^{\prime }=f\left(x\right)g\left(y\right)$$y'=f(x)g(y) Reduces to Separation of Variables

Let $$I$$I be an interval, and let $$J$$J be an interval containing the range of $$y$$y. Suppose that $$f$$f is continuous on $$I$$I and that $$g$$g is continuous and nonzero on $$J$$J. A continuously differentiable function $$y:I\to J$$y:I\to J satisfies

if and only if, for some constant $$C$$C, it satisfies

on $$I$$I.

5.1Proof

Choose antiderivatives $$G$$G and $$F$$F such that $$G^{\prime }\left(y\right)=1/g\left(y\right)$$G'(y)=1/g(y) and $$F^{\prime }\left(x\right)=f\left(x\right)$$F'(x)=f(x). If $$y$$y satisfies the differential equation, then

Hence $$G\left(y\right(x\left)\right)-F\left(x\right)$$G(y(x))-F(x) is constant on $$I$$I, which gives the separated relation. Conversely, differentiating

and applying the chain rule gives

and the hypothesis $$g\left(y\right)\ne 0$$g(y)\ne 0 yields

Constant solutions arising from zeros of $$g\left(y\right)$$g(y) must be checked separately before dividing by $$g\left(y\right)$$g(y).

6Proposition 3: $$y^{\prime }+p\left(x\right)y=q\left(x\right)$$y'+p(x)y=q(x) Reduces to a Product Derivative

Suppose that

on an interval $$I$$I. If a function $$\mu \left(x\right)$$\mu(x) is nonzero on $$I$$I and satisfies

then the original equation is equivalent on $$I$$I to

6.1Proof

Multiplying the original equation by $$\mu$$\mu gives

Since $${\mu }^{\prime }=\mu p\left(x\right)$$\mu'=\mu p(x),

The product rule gives

and hence

Conversely, division by the nonzero function $$\mu$$\mu recovers the original equation, which proves equivalence. The integrating factor is chosen specifically to construct a product derivative.

7Criterion for Exact Differential Form

For an equation of the form

the objective is not to separate $$x$$x and $$y$$y. Instead, determine whether there is a function $$F(x,y)$$F(x,y) such that

If $$F_x=M$$F_x=M and $$F_y=N$$F_y=N, then the equation is $$dF=0$$dF=0, and its solutions are represented implicitly by

Exact equations require partial derivatives and total differentials, so their detailed treatment is deferred to the dedicated lecture.

8A Common Misclassification

The equation

contains $$x$$x and $$y$$y additively. It therefore cannot be separated as

Instead, it can be written as

so it is a first-order linear equation. Classification depends on reducibility to a standard form, not on superficial simplicity.

9Subsequent Learning Sequence

This page has established the reasons for the classification. The following lectures develop the calculations and qualifications for each method.

10Exercise