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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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
a
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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