Standard Form of a First-Order Differential Equation

Every technique in this chapter starts from one of two ways of writing a first-order equation: the standard form y' = f(x,y), which isolates the derivative, and the differential form $M (x, y) d x + N (x, y) d y = 0$ , which treats $x$ and $y$ symmetrically. Moving fluently between them is the first skill to acquire.

Quick Reference

Form	Equation	When it is useful
Most general form	F(x, y, y') = 0	Rarely used directly; the derivative may be tangled
Standard form	$\frac{d y}{d x} = f (x, y)$	Direction fields, linear equations, existence theorems, numerical methods
Differential form	$M (x, y) d x + N (x, y) d y = 0$	Separable equations, exact equations, integrating factors
Converting to differential form	$f (x, y) = - \frac{M (x, y)}{N (x, y)}$	Multiply the standard form through by $d x$
Converting to standard form	$\frac{d y}{d x} = - \frac{M (x, y)}{N (x, y)}$	Divide the differential form by $N d x$

The Standard Form

In this chapter we study first-order differential equations. The standard form of a first-order differential equation is

\frac{dy}{dx} = f(x,y) \tag{A}

where $y$ is the unknown function of $x$ and $f$ is a given function defined on a region $D$ in the $x y$ -plane.

Although form (A) does not cover all first-order differential equations, it is inclusive enough for almost all applications. The most general form for a first-order differential equation is

F(x, y, y') = 0 .

If we assume we can solve for y', which is not always possible, we obtain y' = f(x,y), that is, form (A).

The region $D$ matters. The function $f$ needs to be defined at a point for the equation to say anything there, so the geometry of $D$ controls where solutions can live. In y' = \dfrac{1}{x - y}, for instance, $f$ is undefined along the line $y = x$ , and no solution curve can cross it.

The Differential Form

If $f (x, y)$ is written as a quotient of two functions $M (x, y)$ and $N (x, y)$ , namely

f (x, y) = - \frac{M (x, y)}{N (x, y)},

then form (A) becomes $\frac{d y}{d x} = - \frac{M (x, y)}{N (x, y)}$ , which is equivalent to the differential form

M(x,y)\,dx + N(x,y)\,dy = 0 . \tag{B}

The minus sign in the definition of $f$ is a convention chosen precisely so that form (B) comes out with a plus sign between the two terms.

If we write a differential equation in the form (B), either $x$ or $y$ can be regarded as the independent variable. This symmetry is the practical advantage of the differential form, and it is used constantly in the sections on separable and exact equations.

That freedom is worth more than it might appear. An equation that looks hopeless as y' = f(x,y) sometimes becomes routine when you solve for $x$ as a function of $y$ instead, and the differential form lets you make that switch without rewriting anything.

Converting Between the Forms

Going from standard to differential form takes one step: multiply through by $d x$ .

\frac{d y}{d x} = f (x, y) ⟹ d y = f (x, y) d x ⟹ - f (x, y) d x + d y = 0

so $M = - f (x, y)$ and $N = 1$ .

Going the other way, divide by $N d x$ :

M d x + N d y = 0 ⟹ M + N \frac{d y}{d x} = 0 ⟹ \frac{d y}{d x} = - \frac{M}{N} (N \neq 0)

Write $\frac{d y}{d x} = \frac{x^{2} - y}{2 x y}$ in differential form.

Solution

Multiply both sides by $2 x y d x$ :

2 x y d y = (x^{2} - y) d x .

Move everything to the left:

- (x^{2} - y) d x + 2 x y d y = 0 or (y - x^{2}) d x + 2 x y d y = 0 .

Here $M = y - x^{2}$ and $N = 2 x y$ .

Note that the choice of $M$ and $N$ is not unique. Multiplying the whole equation by any nonzero function gives a different pair $M$ and $N$ describing the same solution curves. That freedom is exactly what integrating factors exploit in Section 2.5.

Write $x^{2} y d x + x^{3} y^{2} d y = 0$ in standard form, and note where the conversion fails.

Solution

Divide by $x^{3} y^{2} d x$ :

\frac{d y}{d x} = - \frac{x^{2} y}{x^{3} y^{2}} = - \frac{1}{x y} .

The conversion required dividing by $x^{3} y^{2}$ , so it is valid only where $x \neq 0$ and $y \neq 0$ . In particular, the constant function $y = 0$ satisfies the original differential form (both terms vanish) but does not appear in the standard form at all, where the right-hand side is undefined. Dividing can lose solutions, so it is worth noting what you divided by.

Show that \left(x y' - y\right)^2 = (x+y)^2 is not in standard form, and put it there.

Solution

The derivative is buried inside a square, so the equation has the general shape F(x,y,y') = 0 rather than y' = f(x,y). Taking the square root of both sides gives

x y' - y = \pm(x+y) ,

which splits into two separate equations:

y' = \frac{y + x + y}{x} = \frac{x + 2y}{x} \qquad \text{and} \qquad y' = \frac{y - x - y}{x} = -1 .

Each branch is now in standard form and can be solved on its own. This is typical: an equation of higher degree in y' usually splits into several standard-form equations, and the full solution set is the union of what each branch produces.

Exercises

Convert $\frac{d y}{d x} = \frac{y - 3 x}{x + y}$ to differential form.

Solution

Multiplying by $(x + y) d x$ gives $(x + y) d y = (y - 3 x) d x$ , so

(3 x - y) d x + (x + y) d y = 0,

with $M = 3 x - y$ and $N = x + y$ .

Convert $(e^{x} + y) d x + (x - \sin y) d y = 0$ to standard form, and also solve it for $d x / d y$ .

Solution

Treating $y$ as the dependent variable,

\frac{d y}{d x} = - \frac{e^{x} + y}{x - \sin y} (x \neq \sin y) .

Treating $x$ as the dependent variable instead,

\frac{d x}{d y} = - \frac{x - \sin y}{e^{x} + y} (e^{x} \neq - y) .

Both describe the same curves. This is the symmetry of the differential form in action: nothing in the original equation distinguishes $x$ from $y$ .

Frequently Asked Questions

Why bother with the differential form at all?

Because it is symmetric in $x$ and $y$ . Two of the most important methods in this chapter, separation of variables and exactness, are stated most naturally in that form. Separation asks whether $M$ depends only on $x$ and $N$ only on $y$ ; exactness compares $\partial M / \partial y$ with $\partial N / \partial x$ . Neither question is easy to even ask in the form y' = f(x,y).

Are

M

and

N

uniquely determined by the equation?

No. Multiplying $M d x + N d y = 0$ by any nonzero function $μ (x, y)$ produces a different pair with the same solution curves. Far from being a defect, this is the whole idea behind integrating factors: you choose $μ$ to make the new pair satisfy the exactness test.

What happens where

N (x, y) = 0

The standard form y' = -M/N breaks down there, which usually signals a vertical tangent: the solution curve is still perfectly well behaved, but $y$ is not a function of $x$ near that point. Switching to $\frac{d x}{d y} = - \frac{N}{M}$ often repairs the description, and this is one of the practical payoffs of the differential form.

Can an equation be first order and still not be of the form y' = f(x,y)?

Yes. Anything of degree higher than one in y', such as \left(y'\right)^2 = 4y, is first order but not in standard form. Solving for y' gives y' = \pm 2\sqrt{y}, two standard-form equations. Equations like this are also where singular solutions appear, since here $y = 0$ solves the original equation but belongs to neither branch's general solution.

Introduction to First-Order Equations

Quick Reference

The Standard Form

The Differential Form

Converting Between the Forms

Exercises

Frequently Asked Questions

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