1. Ch 2.1: Linear Equations;
Method of Integrating Factors
A linear first order ODE has the general form
where f is linear in y. Examples include equations with
constant coefficients, such as those in Chapter 1,
or equations with variable coefficients:
),( ytf
dt
dy
=
)()( tgytp
dt
dy
=+
bayy +−=′

2. Constant Coefficient Case
For a first order linear equation with constant coefficients,
recall that we can use methods of calculus to solve:
Cat
ekkeaby
Ctaaby
dta
aby
dy
a
aby
dtdy
±=+=
+−=−
−=
−
−=
−
∫∫
,/
/ln
/
/
/
,bayy +−=′

3. Variable Coefficient Case:
Method of Integrating Factors
We next consider linear first order ODEs with variable
coefficients:
The method of integrating factors involves multiplying this
equation by a function µ(t), chosen so that the resulting
equation is easily integrated.
)()( tgytp
dt
dy
=+

4. Example 1: Integrating Factor (1 of 2)
Consider the following equation:
Multiplying both sides by µ(t), we obtain
We will choose µ(t) so that left side is derivative of known
quantity. Consider the following, and recall product rule:
Choose µ(t) so that
2/
2 t
eyy =+′
[ ] y
dt
td
dt
dy
tyt
dt
d )(
)()(
µ
µµ +=
t
ettt 2
)()(2)( =⇒=′ µµµ
)()(2)( 2/
teyt
dt
dy
t t
µµµ =+

5. Example 1: General Solution (2 of 2)
With µ(t) = e2t
, we solve the original equation as follows:
[ ]
tt
tt
tt
ttt
t
t
Ceey
Ceye
eye
dt
d
eye
dt
dy
e
etyt
dt
dy
t
eyy
22/
2/52
2/52
2/522
2/
2/
5
2
5
2
2
)()(2)(
2
−
+=
+=
=
=+
=+
=+′
µµµ

6. Method of Integrating Factors:
Variable Right Side
In general, for variable right side g(t), the solution can be
found as follows:
[ ]
atatat
atat
atat
atatat
Cedttgeey
dttgeye
tgeye
dt
d
tgeyae
dt
dy
e
tgtyta
dt
dy
t
tgayy
−−
+=
=
=
=+
=+
=+′
∫
∫
)(
)(
)(
)(
)()()()(
)(
µµµ

7. Example 2: General Solution (1 of 2)
We can solve the following equation
using the formula derived on the previous slide:
Integrating by parts,
Thus
tyy −=+′ 5
5
1
5/5/5/
)5()( tttatatat
CedtteeCedttgeey −−−−
+−=+= ∫∫
[ ]
5/5/
5/5/5/
5/5/5/
550
5525
5)5(
tt
ttt
ttt
tee
dtetee
dttedtedtte
−=
−−=
−=−
∫
∫∫∫
( ) 5/5/5/5/5/
550550 ttttt
CetCeteeey −−−
+−=+−=

8. Example 2: Graphs of Solutions (2 of 2)
The graph on left shows direction field along with several
integral curves.
The graph on right shows several solutions, and a particular
solution (in red) whose graph contains the point (0,50).
5/
5505
5
1 t
Cetytyy −
+−=⇒−+−=′

9. Example 3: General Solution (1 of 2)
We can solve the following equation
using the formula derived on previous slide:
Integrating by parts,
Thus
tyy −=−′ 5
5
1
5/5/5/
)5()( tttatatat
CedtteeCedttgeey +−=+= ∫∫
−−−
[ ]
5/
5/5/5/
5/5/5/
5
5525
5)5(
t
ttt
ttt
te
dtetee
dttedtedtte
−
−−−
−−−
=
+−−−=
−=−
∫
∫∫∫
[ ] 5/5/5/5/
55 tttt
CetCeteey +=+= −

10. Example 3: Graphs of Solutions (2 of 2)
The graph on left shows direction field along with several
integral curves.
The graph on right shows several integral curves, and a
particular solution (in red) whose initial point on y-axis
separates solutions that grow large positively from those that
grow large negatively as t → ∞.
5/
555/ t
Cetytyy +=⇒−=−′

11. Method of Integrating Factors for
General First Order Linear Equation
Next, we consider the general first order linear equation
Multiplying both sides by µ(t), we obtain
Next, we want µ(t) such that µ'(t) = p(t)µ(t), from which it
will follow that
)()( tgytpy =+′
[ ] yttp
dt
dy
tyt
dt
d
)()()()( µµµ +=
)()()()()( ttgyttp
dt
dy
t µµµ =+

12. Integrating Factor for
General First Order Linear Equation
Thus we want to choose µ(t) such that µ'(t) = p(t)µ(t).
Assuming µ(t) > 0, it follows that
Choosing k = 0, we then have
and note µ(t) > 0 as desired.
ktdtpttdtp
t
td
+=⇒= ∫∫∫ )()(ln)(
)(
)(
µ
µ
µ
,)( )( tdtp
et ∫
=µ

13. Solution for
General First Order Linear Equation
Thus we have the following:
Then
tdtp
ettgtyttp
dt
dy
t
tgytpy
∫
==+
=+′
)(
)(where),()()()()(
)()(
µµµµ
[ ]
tdtp
et
t
cdttgt
y
cdttgtyt
tgtyt
dt
d
∫
=
+
=
+=
=
∫
∫
)(
)(where,
)(
)()(
)()()(
)()()(
µ
µ
µ
µµ
µµ

14. Example 4: General Solution (1 of 3)
To solve the initial value problem
first put into standard form:
Then
and hence
( ) ,21,52 2
==−′ ytyyt
0for,5
2
≠=−′ tty
t
y
222
2
2
ln5
5
1
5
1
)(
)()(
CtttCdt
t
t
t
Ctdt
t
t
Cdttgt
y +=



+=
+
=
+
= ∫
∫∫
µ
µ
2
1
ln
ln2
2
)( 1
)(
2
t
eeeet ttdt
t
dttp
===
∫
=∫=






−−
µ

15. Example 4: Particular Solution (2 of 3)
Using the initial condition y(1) = 2 and general solution
it follows that
or equivalently,
22
2ln52)1( tttyCy +=⇒==
,ln5 22
Cttty +=
( )5/2ln5 2
+= tty

16. Example 4: Graphs of Solution (3 of 3)
The graphs below show several integral curves for the
differential equation, and a particular solution (in red)
whose graph contains the initial point (1,2).
( )
22
22
2
2ln5:SolutionParticular
ln5:SolutionGeneral
21,52:IVP
ttty
Cttty
ytyyt
+=
+=
==−′