Page 24 - Advanced engineering mathematics

4 CHAPTER 1 First-Order Differential Equations

Step 1. For y such that G(y) = 0, write the differential form
1
dy = F(x)dx.
G(y)
In this equation, we say that the variables have been separated.
Step 2. Integrate
1

dy = F(x)dx.
G(y)
Step 3. Attempt to solve the resulting equation for y in terms of x. If this is possible, we have
an explicit solution (as in Examples 1.1 through 1.3). If this is not possible, the solution
is implicitly defined by an equation involving x and y (as in Example 1.4).
Step 4. Following this, go back and check the differential equation for any values of y such that
G(y) = 0. Such values of y were excluded in writing 1/G(y) in step (1) and may lead
to additional solutions beyond those found in step (3). This happens in Example 1.1.

EXAMPLE 1.1
2 −x

To solve y = y e ,firstwrite
dy
2 −x
= y e .
dx
If y = 0, this has the differential form
1 −x
dy = e dx.
y 2
The variables have been separated. Integrate
1

−x
dy = e dx
y 2
or
1 −x
− =−e + k
y
in which k is a constant of integration. Solve for y to get
1
y(x) = .
e −x − k
This is a solution of the differential equation for any number k.
Now go back and examine the assumption y = 0 that was needed to separate the variables.
Observe that y = 0 by itself satisfies the differential equation, hence it provides another solution
(called a singular solution).
In summary, we have the general solution
1
y(x) =
e −x − k
for any number k as well as a singular solution y = 0, which is not contained in the general
solution for any choice of k.

This expression for y(x) is called the general solution of this differential equation because
it contains an arbitrary constant. We obtain particular solutions by making specific choices for
k. In Example 1.1,

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October 14, 2010 14:9 THM/NEIL Page-4 27410_01_ch01_p01-42