1550078481-Ordinary_Differential_Equations__Roberts_

The Initial Value Problem y' = f(x, y); y(c) = d 55

2.3 Solution of Simple First-Order Differential

Equations

Sometimes one can solve the initial value problem y' = J(x, y); y(c) = d by

finding the solution ¢(x, y, C) = 0 of the differential equation y' = J(x, y) and

then determining the value of C which satisfies the initial condition y(c) = d.

Perhaps in calculus, you examined various elementary techniques for finding

the solution of the differential equation y' = f(x, y), such as separating vari-

ables, performing a change of variable, testing for exactness and solving those

equations which are exact, and using integrating factors. These techniques

for solving first-order differential equ ations were all developed and employed

prior to 1800. However, relatively few differential equations can actually be

solved using these techniques. In chapter 1, we showed how to verify that a

given function is the solution of a specific differential equation; however, we

did not discuss how to actu ally find a solution. In this section, we shall show

how to solve the differential equation y' = f(x, y) when j(x, y) has one of the

three forms:

(i) j(x, y) = g(x), (ii) j(x, y) = g(x)/h(y), and (iii) j(x, y) = a(x)y+b(x).

2.3.1 Solution of y' =g(x)

In calculus, you solved many linear differential equ ations of the form

(1) y' = g(x)

by finding the antiderivative of g. Integrating equation (1) symbolically, we
find the general solution to be

(2) y(x) = jx g(t) dt + C

where C is an arbitrary constant. By the existence and uniqueness theorem
for first-order linear initial value problems, this solution exists and is unique
on any interval on which the function g(x) is defined and continuous.

EXAMPLE 1 Solving an Initial Value Problem

by Finding the General Solution

a. Find the general solution of the differential equation

(3) y I

1

(x + 2)"