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Hypothesis Testing – Fundamentals of Engineering Exam Review

January 5, 2012 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin reinforces your understanding of functions and function notation, a topic commonly encountered on the Engineer in Training Exam. He will first start with defining a function and function notation, then discuss how to evaluate a function, discuss domain and range, and wrap it up with a discussion on function composition.

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Higher Order Non Homogeneous Differential Equations – Fundamentals of Engineering Exam Review

January 2, 2012 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a FE Exam Review of Higher Order Nonhomogeneous Differential Equations.

We will begin by defining Higher Order NonHomogeneous Differential Equations and then run through a few example problems.

This FE Exam Review of Higher Order Nonhomogeneous Differential Equations is part of the global subject Mathematics.

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Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a FE Exam Review of Higher Order Nonhomogeneous Differential Equations.

We’ll start off this tutorial discussing the theory and then get in to working a problem.

So let’s start off our FE Exam Review of Higher Order Nonhomogeneous Differential Equations by reviewing what exactly a NonHomogeneous Differential Equation is.

A non-homogeneous second-order linear equation takes the form y double prime plus PT y prime plus Q T Y is equal to G T where the general solution

of this equation would be YT is equal to

YP t plus y h t so this is the general

solution of a higher-order non homey and

homogeneous differential equation now YP

T is known as the particular solution of

the non-homogeneous equation and why HT

is the general solution of the

Associated homogeneous equation now we

know how to find the solution to a

homogeneous equation so let’s focus on

how we can determine the particular

solution of a non-homogeneous

differential equation so let’s determine

or let’s focus on how to find YP t of

the general solution now one process we

can use for solving this is known as the

method of undetermined coefficients so

let’s look at that the method of

undetermined coefficients now this

method is based on a guessing technique

where we will guess the form of YP and

then plug it into the equation to find

it however this method only works under

the following conditions number one the

Associated homogeneous equations has

constant coefficients so what the YH

part of the general solution has

constant coefficients in condition

number two is the non-homogeneous term

GT is this is is a special form so the

non the non-homogeneous term G T is in a

special form and that special form can

be one of two things

GT can be P T e to the Alpha X cosine of

beta X or it could be in the form L T e

to the Alpha X sine of beta X so it

needs to be in one of these two forms to

be able to use this method to solve and

PT and L T are both polynomial functions

now there are two disadvantages to this

method first it will only work for a

fairly small of class of gts as we’ve

defined here the class of GT s for which

this method works however does include

some of the more common functions so we

commonly do see these types of functions

as GT however there are many functions

out there for which the undetermined

coefficients simply won’t work second it

is generally only useful for constant

coefficient differential equations now

this method is quite simple all we need

to do is look at GT and make a guess as

to the form of white the YP leaving the

coefficients undetermined hence the name

of the method next we would plug in the

guess to the differential equation and

see if we can determine the values of

the coefficients if we can determine the

values for the coefficients then we

guessed correctly if we can’t find the

values for the coefficients then we

guessed incorrectly and we just have to

start all over so that’s a lot of talk

let’s go through an example let’s uh

let’s reinforce this through seeing how

it works let’s say we want to solve the

equation y double prime minus 4y minus

12y is equal to three e to the 5t we

look at GT and we see that it is in one

of the forms that we can use so we’ll go

ahead and use the method of the method

of undetermined coefficients now the

point here again is to find the

particular solution however the first

thing that we’re going to do is find the

complimentary homogeneous solution to

this differential equation so recall

that the complete complementary solution

comes from solving this equation without

GCT there with where it’s just zero on

the right side so we can rewrite that as

let’s say we’re dealing with H the

homogeneous equation H prime minus four

prime minus 12y is equal to zero now the

characteristic equation here is R

squared minus four R minus twelve which

can be factored as R minus six times R

plus two so we see that r1 is going to

be equal to six and r2 it’s going to be

equal to

– negative – so we have two real

distinct roots therefore our

complementary homogeneous equation y H T

is going to be equal to C 1 e to the 6t

plus C 2 e to the negative 2t so that’s

our homogeneous equation of our general

solution now let’s find the particular

solution we need to make a guess as to

the form of the particular solution to

this differential equation

since G T is an exponential and we know

that Exponential’s never just appear or

disappear in the differentiation process

it seems that a likely form of the

particular solution would be Y P T some

constant e to the 5 T because e does

just doesn’t disappear does not just

appear either so that’s a good guess to

start with now all that we need to do is

do a couple of derivatives and plug this

into the differential equation and see

if we can determine what a would equal

and if we can do that then our guess was

right so let’s go ahead and do a couple

derivatives and so we got Y double prime

is going to be equal to 25 e to the 5t Y

prime is going to be equal to 5 e to the

5 t and we already know what Y is so

let’s go ahead and just plug these

values into our differential equation we

got 25 a to the e 5 2 raised to the 5t

minus 4 a 4 times 5 a e to the 5 t minus

12a e to the 5 T is equal to 3 e to the

5t combining all these terms on

left side we get negative 7a e to the 5

T is equal to 3 e to the 5 T so things

are looking good it looks like we’re

gonna be able to solve for a and in fact

we are and just isolating a we solve a

is equal to negative 3/7 so we found a

value for the coefficient this means

that we guessed correctly

now our particular solution for this

differential equation now will be Y P T

is equal to negative 3/7 e to the 5 T

that’s our particular solution so all we

need to do is we’ll jump up here we just

need to now combine our homogeneous

solution and our particular solution to

get the overall solution of the equation

so we have negative 3/7 e to the 5t plus

C 1 e to the 6t plus C 2 e to the

negative 2t and there it is that is our

solution using the method of

undetermined coefficients so let’s look

at another example here let’s look at

another example and see what happens

when our guess is initially wrong and

what we can do to address that so just

follow along we’re gonna guess something

and it’s going to be wrong and I’ll show

you how we can adjust quickly on-the-fly

to get the answer correctly let’s find

the particular solution for y double

prime minus 4 y prime minus 12 y is

equal to sine of 12 I’m sorry it’s sine

of 2t that’s 2 sine of 2t now we want to

start off by finding the complementary

homogeneous solution before we proceed

on to determining the particular

solution notice though

this solution will be the same as the

first problem we just went over it’s the

same on the left side so we won’t go

through the mess we’ve remembered we can

just recall it at the end of this

problem now recall in the last problem

that we were dealing with an exponential

function function as G sub G of T so we

guess that our particular solution would

include an exponential as well and we

guessed right now this differential

equation has a sign so let’s start with

our guess as y PT is equal to a sine of

2t and let’s see what happens so

differentiating that a couple times and

plugging it back into our differential

equation we get negative 4 a sine of 2t

minus 4 2 times a cosine of 2t minus 12

a sine of 2t is equal to sine of 2t now

combining all the like terms we get

negative 16 a sine of 2t minus 8 cosine

8 times a cosine 2t is equal to sine of

2t now we need to define a so that we

can get the same function on both sides

of the equal sign we do this by setting

the cosines and sines from each side

against one another so that cosine of 2t

is equal to negative 8 a is equal to

zero because there’s no cosine on the

right side here now there’s a negative 8

cosine on the left side so in this case

has to equal zero now we saw set the

signs against each other we got let’s

see we got for the signs we got negative

16 on the left and we have one on the

right so that means a has to be negative

1 16 so these constants obviously are

not the same so since there’s no cosine

on the right hand side this means that

the coefficient must be zero on that

side however in order for the cosine to

drop out as it must in order for the

guests to satisfy the differential

equation we need to set a equals to zero

but a equal to zero then the sine will

also drop out and that can’t happen

because we have sine on the right sides

so likewise reverse we need a to equal

negative 1/16 to equal the number of

signs on the right side but if we plug

that 8 into our cosine on the right side

then that’s not going to satisfy our

equation for the right side for the

cosine so this means that since there’s

two values for a you don’t have to go

through all that blur ‘bitch that I just

went through but since there’s two

values of a it’s obviously that our

guests was wrong

luckily we kind of worked through the

problem already which shows us to how we

can fix our guests we notice that on the

left side when we differentiate that

we’re going to have a combination of

sines and cosines so let’s start our

next guest as our particular solution as

a cosine of 2t plus B sine of 2t now

plugging this into the differential

equation we can differentiate this and

plug it into our original differentiate

differential equation we get negative 4

a cosine of 2t minus 4b sine

of 2t minus four negative to a sine of

2t plus two B cosine of 2t minus 12 a

cosine of 2t plus B sine of 2t is equal

to sine of 2t so that’s a mega long

equation but let’s let’s simplify that a

little bit we got negative 4a minus 8 b

-12 a cosine of 2t plus negative 4b plus

8 a minus 12 B sine of 2t it’s equal to

sine of 2t and simplifying that just a

little bit more we get negative 16 a

minus 8 B cosine of 2t plus 8 a minus 16

B sine of 2t is equal to sine of 2t so

once again we need to set the

coefficients sine and cosine equal to

each other from the left and right

because each side of this equation needs

to equal to determine what a and B would

be so for cosine we’re going to get

negative 16 a minus 8 B is equal to 0

because there’s negative 16 days cosines

on the left side and negative 8b cosines

on the left side and then there’s zero

on the right so sine of 2t is going to

be 8 a minus 16 B is equal to 1 now we

have two equations and two unknowns so

we can go ahead and solve that and we

end up getting a is equal to 1 over 40

and B is equal to negative 1 over 20 so

we actually found some constants for our

particular solution therefore our guess

was correct and now all we need to do is

define that particular solution let’s

start anew let’s just plug our numbers

into our equation we got 1 over 40

cosine of 2t minus 1 divided by 20 sine

of 2t so that is our particular solution

for this specific higher-order non

homogeneous linear differential equation

so all we need to do now is combine that

with our original homogeneous solution

and we would get the overall solution to

the problem so that’s all I got for you

guys if you have any more questions or

concerns or suggestions go ahead and

send them at the engineer and training

exam comm check out my other videos

there and get on your way to dominating

this exam so we’ll be talking soon guys

take care all right, that concludes our FE Exam Review of Higher Order Nonhomogeneous Differential Equations.

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Derivatives – Fundamentals of Engineering Exam Review

December 29, 2011 by Justin Leave a Comment

Right Click to Download this Video

In this episode of Engineer In Training Exam TV we will reinforce your understanding of Derivatives. We will begin by defining Derivatives and then run through a few example problems.

CLICK HERE to get immediate access to the FREE “EIT PREPARATION BOOT CAMP” and get yourself started on the road to DOMINATING the EIT Exam NOW.

Enjoy!!

Curvature – Fundamentals of Engineering Exam Review

December 28, 2011 by Justin Leave a Comment

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In this episode of Engineer In Training Exam TV we will reinforce your understanding of Curvature. We will begin by defining Curvature and then run through a few example problems.

CLICK HERE to get immediate access to the FREE “EIT PREPARATION BOOT CAMP” and get yourself started on the road to DOMINATING the EIT Exam NOW.

Enjoy!!

Confidence Intervals – Fundamentals of Engineering Exam Review

December 26, 2011 by Justin Leave a Comment

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In this episode of Engineer In Training Exam TV we will reinforce your understanding of Confidence Intervals. We will begin by defining Confidence Intervals and then run through a few example problems.

CLICK HERE to get immediate access to the FREE “EIT PREPARATION BOOT CAMP” and get yourself started on the road to DOMINATING the EIT Exam NOW.

Enjoy!!

Higher Order Homogeneous Differential Equations – Fundamentals of Engineering Exam Review

December 23, 2011 by Justin Leave a Comment

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In this episode of Engineer In Training Exam TV we will reinforce your understanding of Higher Order Homogeneous Differential Equations. We will begin by defining Higher Order Homogeneous Differential Equations and then run through a few example problems.

CLICK HERE to get immediate access to the FREE “EIT PREPARATION BOOT CAMP” and get yourself started on the road to DOMINATING the EIT Exam NOW.

Enjoy!!

Simultaneous Linear Equations – Fundamentals of Engineering Exam Review

December 23, 2011 by Justin Leave a Comment

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In this episode of Engineer In Training Exam TV we will reinforce your understanding of Simultaneous Linear Equations. We will begin by defining Simultaneous Linear Equations and then run through a few example problems.

CLICK HERE to get immediate access to the FREE “EIT PREPARATION BOOT CAMP” and get yourself started on the road to DOMINATING the EIT Exam NOW.

Enjoy!!

Taylor and Maclaurin Series – Fundamentals of Engineering Exam Review

December 23, 2011 by Justin Leave a Comment

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In this episode of Engineer In Training Exam TV we will reinforce your understanding of Taylor and Maclaurin Series. We will begin by defining Taylor and Maclaurin Series and then run through a few example problems.

CLICK HERE to get immediate access to the FREE “EIT PREPARATION BOOT CAMP” and get yourself started on the road to DOMINATING the EIT Exam NOW.

Enjoy!!

Hypothesis Testing – Fundamentals of Engineering Exam Review

Higher Order Non Homogeneous Differential Equations – Fundamentals of Engineering Exam Review

Transcription

Derivatives – Fundamentals of Engineering Exam Review

Curvature – Fundamentals of Engineering Exam Review

Confidence Intervals – Fundamentals of Engineering Exam Review

Higher Order Homogeneous Differential Equations – Fundamentals of Engineering Exam Review

Simultaneous Linear Equations – Fundamentals of Engineering Exam Review

Taylor and Maclaurin Series – Fundamentals of Engineering Exam Review

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