MATH 219Fall 2014

Lecture 2

Content: Linear equations; method of integrating factors. (section 2.1)).

Suggested Problems:

2.1: 2, 3, 6, 10, 11, 15, 17, 30, 31, 33, 34, 37

1 Linear Equations, Integrating Factors

Definition 1.1 A first order ODE is said to be linear if it can be written in theform

dy

dt+ p(t)y = q(t)

for some functions p(t) and q(t).

The reason why the name linear is attached to these equations will be discussedlater. In this lecture we will see how we can solve a first order linear ODE.

The right hand side of this equation is a function of t only, so it can be integratedwith respect to t. If the right hand side were a function of y only, it would be aneasy problem, but this is not the case. Or if the left hand side were the derivative ofsomething which can be explicitly written, that would also let us solve the equationeasily. This is not the case either. But remember that if a and b both depend on tthen

d(ab)

dt= a

db

dt+da

dtb. (1)

If we try to make the right hand side of this equation look like the left hand side ofour ODE, then it is natural to take b = y. But then we would need to take a = 1 andconsequently they would match only when p(t) = 0. Instead, multiply the wholeODE with a function (t), to be determined later.

1

(t)dy

dt+ (t)p(t)y = (t)q(t).

If we try to match the left hand side of this equation with the right hand side of (1)instead then we can take b = y and a = (t). The condition for these to match is

d

dt= p(t).

But this is a separable equation, therefore we know how to solve it. Indeed,d

=

p(t)dt

ln || =p(t)dt

(t) = ep(t)dt

Such a (t) is called an integrating factor for the ODE. Note that we need onlyone integrating factor that does the job, and it is unnecessary to find all possibleintegrating factors. Hence we can choose the constant arising from the indefiniteintegral in the formula for (t) as we wish.

With this choice of (t), we can then proceed to solve the ODE

(t)dy

dt+ (t)p(t)y = (t)q(t)

d((t)y)

dt= (t)q(t)

(t)y =

(t)q(t)dt

y =

(t)q(t)dt

(t)

Example 1.1 Find all solutions of

y y tet = 0.

Solution: Let us first write the equation in the form

y y = tet.

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Therefore the equation is linear with p(t) = 1 and q(t) = tet. As in the discussionabove, set

(t) = ep(t)dt = e

1dt = et.Multiplying the ODE by (t), we get

ety ety = t(ety) = t

ety =tdt =

t2

2+ c

y =t2et

2+ cet

Example 1.2 Find the solution of the differential equation

ty + 3y =2

t

with the initial condition y(1) = 2.

Solution: First, let us divide by t and rewrite the equation as

y +3

ty =

2

t2,

therefore the equation is linear with p(t) = 3t

and q(t) = 2t2

. Set

(t) = ep(t)dt = e

3tdt = e3 ln t = t3.

(The diligent reader may worry about the absolute value in ln after integration. Thisdoes not cause a problem essentially because both t3 are integrating factors. Theequation is problematic at t = 0 anyway, so the intervals (0,) and (, 0) can betreated separately.) Next, multiply the ODE with (t) and we have

t3y + 3t2y = 2t

(t3y) = 2t

t3y = t2 + c.

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In order to determine c, we can use the initial condition at this point:

13y(1) = 12 + c,

therefore c = 1. Hence the solution is

y =1

t+

1

t3.

Example 1.3 Let a > 0. Show that all solutions of the differential equation

y + ay = b

have the same limit as t, given that a > 0 and b are constants.Solution: The integrating factor for this equation is (t) = e

adt = eat. Multiplying

the ODE by (t), we get

eaty + eatay = beat

(eaty) = beat

eaty =

beatdt =

b

aeat + c

y =b

a+ ceat.

Since a > 0, we have limt ceat = 0 for any value of c. Therefore limt

y(t) =b

a,

and this is independent of which solution we pick.

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