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First-Order Linear Differential Equations
A first-order linear differential equation is one that can be put into the form

dy
1 1 Psxdy − Qsxd
dx

where P and Q are continuous functions on a given interval. This type of equation occurs
frequently in various sciences, as we will see.
An example of a linear equation is xy9 1 y − 2x because, for x ± 0, it can be written
in the form

1
2 y9 1 y−2
x
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Notice that this differential equation is not separable because it’s impossible to factor the
expression for y9 as a function of x times a function of y. But we can still solve the equa­tion
by noticing, by the Product Rule, that
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xy9 1 y − sxyd9

and so we can rewrite the equation as

sxyd9 − 2x

If we now integrate both sides of this equation, we get

C
xy − x 2 1 C    or    y − x 1
x

If we had been given the differential equation in the form of Equation 2, we would have had
to take the preliminary step of multiplying each side of the equation by x.
It turns out that every first-order linear differential equation can be solved in a simi-
lar fashion by multiplying both sides of Equation 1 by a suitable function Isxd called an
integrating factor. We try to find I so that the left side of Equation 1, when multiplied by
Isxd, becomes the derivative of the product Isxdy:

3 Isxd( y9 1 Psxdy) − ( Isxdy)9

If we can find such a function I, then Equation 1 becomes

(Isxdy)9 − Isxd Qsxd
Integrating both sides, we would have

Isxdy − y Isxd Qsxd dx 1 C

so the solution would be


4 ysxd −
1
Isxd
Fy G
Isxd Qsxd dx 1 C


To find such an I, we expand Equation 3 and cancel terms:

Isxdy9 1 Isxd Psxdy − sIsxdyd9 − I9sxdy 1 Isxdy9

Isxd Psxd − I9sxd


1

, 2 ■ FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS



This is a separable differential equation for I, which we solve as follows:

dI
y I
− y Psxd dx


ln I − y Psxd dx
||
I − Ae y Psxd dx

where A − 6e C. We are looking for a particular integrating factor, not the most general
one, so we take A − 1 and use

5 Isxd − e y Psxd dx

Thus a formula for the general solution to Equation 1 is provided by Equation 4, where I is
given by Equation 5. Instead of memorizing this formula, however, we just remember the
form of the integrating factor.


To solve the linear differential equation y9 1 Psxdy − Qsxd, multiply both sides by
the integrating factor Isxd − e y Psxd dx and integrate both sides.


dy
EXAMPLE 1 Solve the differential equation 1 3x 2 y − 6x 2.
dx
SOLUTION  The given equation is linear since it has the form of Equation 1 with
Psxd − 3x 2 and Qsxd − 6x 2. An integrating factor is

Isxd − e y 3x
2
dx 3
− ex
Figure 1 shows the graphs of several
3
members of the family of solutions Multiplying both sides of the differential equation by e x , we get
in Example 1. Notice that they all
approach 2 as x l `. dy 3
1 3x 2e x y − 6x 2e x
3 3
ex
6 dx
C=2
C=1
d 3
se x yd − 6x 2e x
3
or
C=0 dx
C=_1
_1.5 1.8 Integrating both sides, we have
C=_2
e x y − y 6x 2e x dx − 2e x 1 C
3 3 3

_3

FIGURE 1 y − 2 1 Ce2x
3
n


EXAMPLE 2 Find the solution of the initial-value problem

x 2 y9 1 xy − 1      x . 0      ys1d − 2

SOLUTION  We must first divide both sides by the coefficient of y9 to put the differential
equation into standard form:

1 1
6 y9 1 y − 2     x . 0
x x