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Chapter 1 Summary

In this chapter we have discussed applications of and solution methods for several important types of first-order differential equations, including those that are separable (Section 1.4), linear (Section 1.5), or exact (Section 1.6). In Section 1.6 we also discussed substitution techniques that can sometimes be used to transform a given first-order differential equation into one that is either separable, linear, or exact.

Lest it appear that these methods constitute a “grab bag” of special and unrelated techniques, it is important to note that they are all versions of a single idea. Given a differential equation

f (x, y, y ′) = 0,

$f (x, y, y ′) = 0,$ (1)

we attempt to write it in the form

\frac{d}{d x} [G (x, y)] = 0 .

$\frac{d}{d x} [G (x, y)] = 0 .$ (2)

It is precisely to obtain the form in Eq. (2) that we multiply the terms in Eq. (1) by an appropriate integrating factor (even if all we are doing is separating the variables). But once we have found a function $G (x, y)$ $G (x, y)$ such that Eqs. (1) and (2) are equivalent, a general solution is defined implicitly by means of the equation

G (x, y) = C

$G (x, y) = C$ (3)

that one obtains by integrating Eq. (2).

Given a specific first-order differential equation to be solved, we can attack it by means of the following steps:

Is it separable? If so, separate the variables and integrate (Section 1.4).
Is it linear? That is, can it be written in the form

$\frac{d y}{d x} + P (x) y = Q (x) ?$ $\frac{d y}{d x} + P (x) y = Q (x) ?$

If so, multiply by the integrating factor $ρ = \exp (\int P d x)$ $ρ = \exp (\int P d x)$ of Section 1.5.
Is it exact? That is, when the equation is written in the form $M d x + N d y = 0$ $M d x + N d y = 0$ , is $\partial M / \partial y = \partial N / \partial x$ $\partial M / \partial y = \partial N / \partial x$ (Section 1.6)?
If the equation as it stands is not separable, linear, or exact, is there a plausible substitution that will make it so? For instance, is it homogeneous (Section 1.6)?

Many first-order differential equations succumb to the line of attack outlined here. Nevertheless, many more do not. Because of the wide availability of computers, numerical techniques are commonly used to approximate the solutions of differential equations that cannot be solved readily or explicitly by the methods of this chapter. Indeed, most of the solution curves shown in figures in this chapter were plotted using numerical approximations rather than exact solutions. Several numerical methods for the appropriate solution of differential equations will be discussed in Chapter 2.

Chapter 1 Review Problems

Find general solutions of the differential equations in Problems 1 through 30. Primes denote derivatives with respect to x.

$x^{3} + 3 y - x y^{′} = 0$ $x^{3} + 3 y - x y^{′} = 0$
$x y^{2} + 3 y^{2} - x^{2} y^{′} = 0$ $x y^{2} + 3 y^{2} - x^{2} y^{′} = 0$
$x y + y^{2} - x^{2} y^{′} = 0$ $x y + y^{2} - x^{2} y^{′} = 0$
$2 x y^{3} + e^{x} + (3 x^{2} y^{2} + \sin y) y^{′} = 0$ $2 x y^{3} + e^{x} + (3 x^{2} y^{2} + \sin y) y^{′} = 0$
$3 y + x^{4} y^{′} = 2 x y$ $3 y + x^{4} y^{′} = 2 x y$
$2 x y^{2} + x^{2} y^{′} = y^{2}$ $2 x y^{2} + x^{2} y^{′} = y^{2}$
$2 x^{2} y + x^{3} y^{′} = 1$ $2 x^{2} y + x^{3} y^{′} = 1$
$2 x y + x^{2} y^{′} = y^{2}$ $2 x y + x^{2} y^{′} = y^{2}$
$x y^{′} + 2 y = 6 x^{2} \sqrt{y}$ $x y^{′} + 2 y = 6 x^{2} \sqrt{y}$
$y^{′} = 1 + x^{2} + y^{2} + x^{2} y^{2}$ $y^{′} = 1 + x^{2} + y^{2} + x^{2} y^{2}$
$x^{2} y^{′} = x y + 3 y^{2}$ $x^{2} y^{′} = x y + 3 y^{2}$
$6 x y^{3} + 2 y^{4} + (9 x^{2} y^{2} + 8 x y^{3}) y^{′} = 0$ $6 x y^{3} + 2 y^{4} + (9 x^{2} y^{2} + 8 x y^{3}) y^{′} = 0$
$4 x y^{2} + y^{′} = 5 x^{4} y^{2}$ $4 x y^{2} + y^{′} = 5 x^{4} y^{2}$
$x^{3} y^{′} = x^{2} y - y^{3}$ $x^{3} y^{′} = x^{2} y - y^{3}$
$y^{′} + 3 y = 3 x^{2} e^{- 3 x}$ $y^{′} + 3 y = 3 x^{2} e^{- 3 x}$
$y ′ = x^{2} - 2 x y + y^{2}$ $y ′ = x^{2} - 2 x y + y^{2}$
$e^{x} + y e^{x y} + (e^{y} + x e^{y x}) y^{′} = 0$ $e^{x} + y e^{x y} + (e^{y} + x e^{y x}) y^{′} = 0$
$2 x^{2} y - x^{3} y^{′} = y^{3}$ $2 x^{2} y - x^{3} y^{′} = y^{3}$
$3 x^{5} y^{2} + x^{3} y^{′} = 2 y^{2}$ $3 x^{5} y^{2} + x^{3} y^{′} = 2 y^{2}$
$x y^{′} + 3 y = 3 x^{- 3/2}$ $x y^{′} + 3 y = 3 x^{- 3/2}$
$(x^{2} - 1) y ′ + (x - 1) y = 1$ $(x^{2} - 1) y ′ + (x - 1) y = 1$
$x y^{′} = 6 y + 12 x^{4} y^{2/3}$ $x y^{′} = 6 y + 12 x^{4} y^{2/3}$
$e^{y} + y \cos x + (x e^{y} + \sin x) y^{′} = 0$ $e^{y} + y \cos x + (x e^{y} + \sin x) y^{′} = 0$
$9 x^{2} y^{2} + x^{3/2} y^{′} = y^{2}$ $9 x^{2} y^{2} + x^{3/2} y^{′} = y^{2}$
$2 y + (x + 1) y^{′} = 3 x + 3$ $2 y + (x + 1) y^{′} = 3 x + 3$
$9 x^{1/2} y^{4/3} - 12 x^{1/5} y^{3/2} + (8 x^{3/2} y^{1/3} - 15 x^{6/5} y^{1/2}) y^{′} = 0$ $9 x^{1/2} y^{4/3} - 12 x^{1/5} y^{3/2} + (8 x^{3/2} y^{1/3} - 15 x^{6/5} y^{1/2}) y^{′} = 0$
$3 y + x^{3} y^{4} + 3 x y^{′} = 0$ $3 y + x^{3} y^{4} + 3 x y^{′} = 0$
$y + x y^{′} = 2 e^{2 x}$ $y + x y^{′} = 2 e^{2 x}$
$(2 x + 1) y^{′} + y = {(2 x + 1)}^{3/2}$ $(2 x + 1) y^{′} + y = {(2 x + 1)}^{3/2}$
$y^{′} = \sqrt{x + y}$ $y^{′} = \sqrt{x + y}$

Each of the differential equations in Problems 31 through 36 is of two different types considered in this chapter—separable, linear, homogeneous, Bernoulli, exact, etc. Hence, derive general solutions for each of these equations in two different ways; then reconcile your results.

$\frac{d y}{d x} = 3 (y + 7) x^{2}$ $\frac{d y}{d x} = 3 (y + 7) x^{2}$
$\frac{d y}{d x} = {x y}^{3} - x y$ $\frac{d y}{d x} = {x y}^{3} - x y$
$\frac{d y}{d x} = - \frac{3 x^{2} + 2 y^{2}}{4 x y}$ $\frac{d y}{d x} = - \frac{3 x^{2} + 2 y^{2}}{4 x y}$
$\frac{d y}{d x} = \frac{x + 3 y}{y - 3 x}$ $\frac{d y}{d x} = \frac{x + 3 y}{y - 3 x}$
$\frac{d y}{d x} = \frac{2 x y + 2 x}{x^{2} + 1}$ $\frac{d y}{d x} = \frac{2 x y + 2 x}{x^{2} + 1}$
$\frac{d y}{d x} = \frac{\sqrt{y} - y}{\tan x}$ $\frac{d y}{d x} = \frac{\sqrt{y} - y}{\tan x}$

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Table of Contents for Chapter 1 Summary

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Table of Contents for
Chapter 1 Summary