Boyce/DiPrima 9th ed, Ch 2.1: Linear Equations; Method of Integrating FactorsElementary Differential Equations and Boundary Value Problems, 9th edition, by William E. Boyce and Richard C. DiPrima, ©2009 by John Wiley & Sons, Inc.

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:

),( ytfdt

dy

)()( tgytpdt

dy

bayy

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

dtaaby

dy

aaby

dtdy

,/

/ln

/

/

/

,bayy

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 (t), chosen so that the resulting equation is easily integrated.

)()( tgytpdt

dy

Example 1: Integrating Factor (1 of 2)

Consider the following equation:

Multiplying both sides by (t), we obtain

We will choose (t) so that left side is derivative of known quantity. Consider the following, and recall product rule:

Choose (t) so that

3/

21

21 teyy

ydt

td

dt

dytyt

dt

d )()()(

2/)()()(2

1 tettt

)()()( 3/

2

1

2

1teyt

dt

dyt t

Example 1: General Solution (2 of 2)

With (t) = et/2, we solve the original equation as follows:

:solution general

2/3/

6/52/

6/52/

6/52/2/

3/

5

3

5

3

2

1

2

1

2

1

2

1

2

1

tt

tt

tt

ttt

t

Ceey

Ceye

eyedt

d

eyedt

dye

eyy

1 2 3 4 5 6t

1

1

2

3

yt2/3/

5

3 :Solutions Sample tt Ceey

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

tgeyedt

d

tgeyaedt

dye

tgtytadt

dyt

tgayy

)(

)(

)(

)(

)()()()(

)(

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 42

tttatatat CedtteeCedttgeey 222 )4()(

tt

ttt

ttt

tee

dtetee

dttedtedtte

22

225/

222

2

1

2

1

2

1

4

7

2

4)4(

ttttt CetCeteeey 22222

2

1

2

1

4

7

4

7

Example 2: Graphs of Solutions (2 of 2)

The graph shows the direction field along with several integral curves. If we set C = 0, the exponential term drops out and you should notice how the solution in that case, through the point (0, -7/4), separates the solutions into those that grow exponentially in the positive direction from those that grow exponentially in the negative direction..

tCety 2

2

1

4

7 0 .5 1 .0 1 .5 2 .0

t

4

3

2

1

0

yt

tyy 42

Method of Integrating Factors for General First Order Linear Equation

Next, we consider the general first order linear equation

Multiplying both sides by (t), we obtain

Next, we want (t) such that '(t) = p(t)(t), from which it will follow that

)()( tgytpy

yttpdt

dytyt

dt

d)()()()(

)()()()()( ttgyttpdt

dyt

Integrating Factor for General First Order Linear Equation

Thus we want to choose (t) such that '(t) = p(t)(t).

Assuming (t) > 0, it follows that

Choosing k = 0, we then have

and note (t) > 0 as desired.

ktdtpttdtpt

td )()(ln)(

)(

)(

,)( )( tdtpet

Solution forGeneral First Order Linear Equation

Thus we have the following:

Then

tdtpettgtyttpdt

dyt

tgytpy

)()( where),()()()()(

)()(

tdtpett

cdttgty

cdttgtyt

tgtytdt

d

)()( where,)(

)()(

)()()(

)()()(

Example 3: General Solution (1 of 2)

To solve the initial value problem

first put into standard form:

Then

and hence

,21,42 2 ytyyt

0for ,42

ttyt

y

2

2322

2

41)4(

)(

)()(

t

CtCdtt

tt

Cdttt

t

Cdttgty

2lnln22

)( 2

)( teeeet ttdtt

dttp

Example 3: Particular Solution (2 of 2)

Using the initial condition y(1) = 2 and general solution

it follows that

The graphs below show solution curves for the differential equation, including a particular solution whose graph contains the initial point (1,2). Notice that when C=0, we get the parabolic solution (shown)and that solution separ-ates the solutions intothose that are asymptoticto the positive versusnegative y-axis.

121)1(,2

2 CCyt

Cty

22 1

tty

,21,42 2 ytyyt

2 1 1 2t

2

1

1

2

3

4

5t y

22

t

Cty

(1,2)

2ty

Example 4: A Solution in Integral Form (1 of 2)

To solve the initial value problem

first put into standard form:

Then

and hence

,10,22 ytyy

12

yt

y

42)(

2

)(t

dtt

dttpeeet

4/

0

4/4/

0

4/4/ 22222 tt stt st CedseeCdseey

Example 4: A Solution in Integral Form (2 of 2)

Notice that this solution must be left in the form of an integral, since there is no closed form for the integral.

Using software such as Mathematica or Maple, we can approximate the solution for the given initial conditions as well as for other initial conditions.

Several solution curves are shown.

1 2 3 4 5 6t

3

2

1

1

2

3

yt

4/

0

4/4/ 222 tt st Cedseey

,10,22 ytyy

4/

0

4/4/ 222 tt st Cedseey