1550078481-Ordinary_Differential_Equations__Roberts_

The Initial Value Problem y' = f(x, y); y(c) = d 59

variables." Once an equation is in the form (9), the process of separating vari-

ables is completed by replacing y' by dy/ dx and then multiplying the resulting

equation by h(y) dx to obtain

(10) h(y) dy = g(x) dx.

In this form, you can clearly see that the variables x and y are "separated"

and associated with their respective differentials. Symbolically integrating

equation (10) yields

(11) j h(y) dy = j g(x) dx.

The expli cit solution of the separable differential equation (9) and the deter-

mination of the interval of existence of the solution depends upon being able

to represent both integrals appearing in equation (11) as elementary functions

and upon being able to solve the resulting equation for y.

EXAMPLE 2 Solving a Separable Differential Equation

Solve the differential equation

(12) y' = -x/y.

SOLUTION

The direction field for the differential equation y' = -x/y appears in Fig-

ure 2.9. The equation y' = -x/y is already in the form of a separable differ-
ential equation. Replacing y' by dy / dx and multiplying t he resulting equation
by y dx, we get

Integration yields

or

(13)

ydy = -xdx.

j ydy = j -xdx

y2 - x2

-=-+C

2 2 '

where C is an arbitrary constant.

(NOTE: When integrating a separable equation, there is no need for two con-

stants of integration- one constant, c 1 , for the left-hand side of the equation

and another constant, c 2 , for the right-hand side of the equation, s ince these