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Differential Equations

2007 Paul Dawkins 1 http://tutorial.math.lamar.edu/terms.aspx

Preface

Here are my online notes for my differential equations course that I teach here at LamarUniversity. Despite the fact that these are my class notes they should be accessible to anyone

wanting to learn how to solve differential equations or needing a refresher on differential

equations.

Ive tried to make these notes as self contained as possible and so all the information needed toread through them is either from a Calculus or Algebra class or contained in other sections of thenotes.

A couple of warnings to my students who may be here to get a copy of what happened on a daythat you missed.

1.

Because I wanted to make this a fairly complete set of notes for anyone wanting to learndifferential equations I have included some material that I do not usually have time tocover in class and because this changes from semester to semester it is not noted here.

You will need to find one of your fellow class mates to see if there is something in thesenotes that wasnt covered in class.

2. In general I try to work problems in class that are different from my notes. However,

with Differential Equation many of the problems are difficult to make up on the spur ofthe moment and so in this class my class work will follow these notes fairly close as far

as worked problems go. With that being said I will, on occasion, work problems off thetop of my head when I can to provide more examples than just those in my notes. Also, I

often dont have time in class to work all of the problems in the notes and so you will

find that some sections contain problems that werent worked in class due to timerestrictions.

3.

Sometimes questions in class will lead down paths that are not covered here. I try to

anticipate as many of the questions as possible in writing these up, but the reality is that I

cant anticipate all the questions. Sometimes a very good question gets asked in class

that leads to insights that Ive not included here. You should always talk to someone whowas in class on the day you missed and compare these notes to their notes and see whatthe differences are.

4. This is somewhat related to the previous three items, but is important enough to merit its

own item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!!

Using these notes as a substitute for class is liable to get you in trouble. As already notednot everything in these notes is covered in class and often material or insights not in thesenotes is covered in class.


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2007 Paul Dawkins ii http://tutorial.math.lamar.edu/terms.aspx


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Complex Roots

In this section we will be looking at solutions to the differential equation

0ay by cy + + = in which roots of the characteristic equation,

2

0ar br c+ + = are complex roots in the form 1,2r il m= .

Now, recall that we arrived at the characteristic equation by assuming that all solutions to the

differential equation will be of the form

( ) rty t = e

Plugging our two roots into the general form of the solution gives the following solutions to thedifferential equation.

( ) ( ) ( ) ( )1 2andt i t

y t y tl m l m+ -

= =e e

Now, these two functions are nice enough (theres those words again well get around to

defining them eventually) to form the general solution. We do have a problem however. Sincewe started with only real numbers in our differential equation we would like our solution to onlyinvolve real numbers. The two solutions above are complex and so we would like to get our

hands on a couple of solutions (nice enough of course) that are real.

To do this well need Eulers Formula.

cos sini iq q q= +e

A nice variant of Eulers Formula that well need is.

( ) ( )cos sin cos sini i iq q q q q - = - + - = -e

Now, split up our two solutions into exponentials that only have real exponents and exponentials

that only have imaginary exponents. Then use Eulers formula, or its variant, to rewrite the

second exponential.

( ) ( ) ( )( )

( ) ( ) ( )( )1

2

cos sin

cos sin

t i t t

t i t t

y t t i t

y t t i t

l m l

l m l

m m

m m-

= = +

= = -

e e e

e e e

This doesnt eliminate the complex nature of the solutions, but it does put the two solutions into a

form that we can eliminate the complex parts.

Recall from thebasics sectionthat if two solutions are nice enough then any solution can bewritten as a combination of the two solutions. In other words,

( ) ( ) ( )1 1 2 2y t c y t c y t= + will also be a solution.

Using this lets notice that if we add the two solutions together we will arrive at.

( ) ( ) ( )1 2 2 costy t y t tl m+ = e


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This is a real solution and just to eliminate the extraneous 2 lets divide everything by a 2. This

gives the first real solution that were after.

( ) ( ) ( ) ( )1 21 1

cos2 2

tu t y t y t t l m= + = e

Note that this is just equivalent to taking

1 2

1

2c c= =

Now, we can arrive at a second solution in a similar manner. This time lets subtract the two

original solutions to arrive at.

( ) ( ) ( )1 2 2 sinty t y t i tl m- = e

On the surface this doesnt appear to fix the problem as the solution is still complex. However,upon learning that the two constants, c1and c2can be complex numbers we can arrive at a real

solution by dividing this by 2i. This is equivalent to taking

1 21 1and2 2

c ci i= = -

Our second solution will then be

( ) ( ) ( ) ( )1 21 1

sin2 2

tv t y t y t t i i

l m= - = e

We now have two solutions (well leave it to you to check that they are in fact solutions) to the

differential equation.

( ) ( ) ( ) ( )cos and sint tu t t v t t l lm m= =e e

It also turns out that these two solutions are nice enough to form a general solution.

So, if the roots of the characteristic equation happen to be 1,2r il m= the general solution to

the differential equation is.

( ) ( ) ( )1 2cos sint ty t c t c tl lm m= +e e

Lets take a look at a couple of examples now.

Example 1 Solve the following IVP.

( ) ( )4 9 0 0 0 0 8y y y y y - + = = = -

SolutionThe characteristic equation for this differential equation is.2 4 9 0r r- + =

The roots of this equation are 1,2 2 5r i= . The general solution to the differential equation is

then.

( ) ( ) ( )2 21 2cos 5 sin 5t ty t c t c t= +e e


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Now, youll note that we didnt differentiate this right away as we did in the last section. Thereason for this is simple. While the differentiation is not terribly difficult, it can get a little messy.So, first looking at the initial conditions we can see from the first one that if we just applied it we

would get the following.

( ) 10 0y c= =

In other words, the first term will drop out in order to meet the first condition. This makes the

solution, along with its derivative

( ) ( )( ) ( ) ( )

2

2

2 2

2 2

sin 5

2 sin 5 5 cos 5

t

t t

y t c t

t c t c t

=

= +

e

e e

A much nicer derivative than if wed done the original solution. Now, apply the second initialcondition to the derivative to get.

( ) 2 28

8 0 55

y c c- = = = -

The actual solution is then.

( ) ( )28 sin 55

tt t= - e


( ) ( )8 17 0 0 4 0 1y y y y y - + = = - = - SolutionThe characteristic equation this time is.

2 8 17 0r r- + =

The roots of this are 1,2 4r i= . The general solution as well as its derivative is

( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

4 4

1 2

4 4 4 4

1 1 2 2

cos sin

4 cos sin 4 sin cos

t t

t t t t

y t c t c t

y t c t c t c t c t

= +

= - + +

e e

e e e e

Notice that this time we will need the derivative from the start as we wont be having one of the

terms drop out. Applying the initial conditions gives the following system.

( )

( )

1

1 2

4 0

1 0 4

y c

y c c

- = =

- = = +

Solving this system gives 1 4c = - and 2 15c = . The actual solution to the IVP is then.

( ) ( ) ( )4 44 cos 15 sint ty t t t= - +e e


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( ) ( )4 24 37 0 1 0y y y y yp p + + = = = SolutionThe characteristic equation this time is.

24 24 37 0r r+ + =

The roots of this are 11,2 23r i= - . The general solution as well as its derivative is

( )

( )

3 3

1 2

3 3 3 31 21 2

cos sin2 2

3 cos sin 3 sin cos2 2 2 2 2 2

t t

t t t t

t ty t c c

c ct t t t y t c c

- -

- - - -

= +

= - - - +

e e

e e e e

Applying the initial conditions gives the following system.

( )

( )

3 3 3

1 2 2

3 312

1 cos sin

2 2

0 32

y c c c

cy c

p p p

p p

p pp

p

- - -

- -

= = + =

= = - -

e e e

e e

Do not forget to plug the t = pinto the exponential! This is one of the more common mistakesthat students make on these problems. Also, make sure that you evaluate the trig functions as

much as possible in these cases. It will only make your life simpler. Solving this system gives3 3

1 26c cp p= - =e e

The actual solution to the IVP is then.

( )

( ) ( ) ( )

3 3 3 3

3 3

6 cos sin2 2

6 cos sin2 2

t t

t t

t t

y t

t ty t

p p

p p

- -

- - - -

= - +

= - +

e e e e

e e

Lets do one final example before moving on to the next topic.


16 0 10 32 2

y y y yp p + = = - =

Solution

The characteristic equation for this differential equation and its roots are.2 16 0 4r r i+ = =

The general solution to this differential equation and its derivative is.

( ) ( ) ( )

( ) ( ) ( )

1 2

1 2

cos 4 sin 4

4 sin 4 4 cos 4

y t c t c t

y t c t c t

= +

= - +


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Plugging in the initial conditions gives the following system.

1 1

2 2

10 102

33 4

2 4

y c c

y c c

p

p

- = = = -

= = =

So, the constants drop right out with this system and the actual solution is.

( ) ( ) ( )3

10cos 4 sin 44

y t t t= - +