Question

Does anyone know how to solve this: y′′ − 2y′ − 3y = cos(2x)

Answer 1

yes

Answer 2

determine the homogenous part, then that can be used to determine the rest of it with either a wronskian approach or a parameters approach. They even start you off in this subject with a "table" approach

Answer 3

Thank you. My lecturer did not really provide any good lecture material. So it's a bit hard to understand the mathematical terms that you used lol

Answer 4

can you solve a quadratic equation?

Answer 5

yes

Answer 6

some background then: assume e^(rx) is a solution

y = e^(rx)
y' = r e^(rx)
y'' = r^2 e^(rx)

sub it in for the homogenous part

y′′ − 2y′ − 3y = 0

r^2 e^(rx) -2r e^(rx) -3e^(rx) = 0


factor out the e^(rx)

e^(rx) (r^2 -2r -3) = 0 solve the quadratic part of it for r

Answer 7

Hey thanks. I would be able to solve it if the cos(2x) wasn't there. Don't know what to do with the cos(2x).

Answer 8

parameters is the long version, the wronskian is the shortcut. but its good to use parameters to understand why the wronskian is working

Answer 9

im most likely confusing undetermined coeffs with parameters ... just too many words to keep track of lol

Answer 10

Oh okay. I'm really sorry. I'm probably annoying you, but Ive never heard of anything called wronskian

Answer 11

its a guys name, its related to the Cramer rule, another guys name. Its a matrix solution once you know the homogenous parts,

Answer 12

Oh right. Hmm so do I have to find out how to solve this problem using wronskian?

Answer 13

its folded away neatly into some randomized corner of me brain .... yes

Answer 14

but lets solve the homogenous part; what do you get for "r" ?

Answer 15

r=3 and r=-1

Answer 16

then: y1 = A e^(3x), y2= B 2^(-x)

y = yh + yp

yh = y1+y2
= A e^(3x) + B e^(-x) ; is part of the solution

yp is some function of x that we will now attempt to determine

Answer 17

assume A and B are functions of x

yp = A e^(3x) + B e^(-x)
yp' = 3A e^(3x) - B e^(-x) + A' e^(3x) + B' e^(-x) , let A' and B' parts = 0
to account for the
homogenous stuff
yp'' = 9A e^(3x) + B e^(-x) + 3A' e^(3x) - B' e^(-x)

now we can apply this into our setup

Answer 18

y′′ − 2y′ − 3y = cos(2x)

-3yp = -3A e^(3x) -3B e^(-x)
-2yp' = -6A e^(3x) + 2B e^(-x)
yp'' = 9A e^(3x) + B e^(-x) + 3A' e^(3x) - B' e^(-x)
-------------------------------------------------
cos(2x) = 0 + 0 + 3A' e^(3x) - B' e^(-x)

Answer 19

we now have 2 equations in 2 unknowns

A' e^(3x) + B' e^(-x) = 0
A' 3e^(3x) - B' e^(-x) = cos(2x)

solve for A' and B'

Answer 20

Wow this is so long. I need to go through it again to make sure I understand it. Thank you so much!!! =D

Answer 21

learn something new every day!
or try to

Answer 22

Yeah lol. Maths isn't easy at all!! I will do lots of research after my exams

Answer 23

the Wronskian just assumes that you have already done a hundred of these and are at the end :)\[\begin{vmatrix} W_x&W&W_y\\
e^{3x}&0&e^{-x}\\
3e^{3x}&cos(2x)&-e^{-x}\end{vmatrix}\]
the parts needed to apply the Wronskian is the "determinant" of the matrix solution to this cramerized setup

Answer 24

Oh that's quite interesting. I will look into that in more detail. Thanks once again!!

Answer 25

youre welcome, if you havent come across this stuff yet, then its a good primer for whats in the next few chapters lol

Answer 26

Lol thanks for the advice. Will prepare myself for next year then :)

Answer 27

I did not get the right answer =(

Answer 28

\[Ae ^{-x}+Be ^{3x}-\frac{ 1 }{ 65 }[7\cos(2x)+4\sin(2x)]\]
this is what I should get "/

Answer 29

A' e^(3x) + B' e^(-x) = 0
A' 3e^(3x) - B' e^(-x) = cos(2x)
----------------------------
A' 4e^(3x) = cos(2x)

A' = cos(2x) e^(-3x)
\[A=\int cos(2x) e^{-3x}=\frac1{13}e^{-3x}(2sin(2x)-3cos(2x))\]

.................................................................

A' e^(3x) + B' e^(-x) = 0
-A' 3e^(3x) + B' e^(-x) = -cos(2x)
----------------------------
B' 2e^(-x) = -cos(2x)

B' = -2cos(2x) e^(x)
\[B=\int -2cos(2x) e^{x}=-\frac25 e^{x}(2sin(2x)+3cos(2x))\]
\[y=ae^{3x}+be^{-x}+\frac1{13}e^{-3x}e^{3x}(2sin(2x)-3cos(2x))-\frac25 e^{x}e^{-x}(2sin(2x)+3cos(2x))\]
\[y=ae^{3x}+be^{-x}+\frac1{13}(2sin(2x)-3cos(2x))-\frac25 (2sin(2x)+3cos(2x))\]
\[y=ae^{3x}+be^{-x}+\frac2{13}sin(2x)-cos(2x)-\frac45sin(2x)-\frac65cos(2x)\]
\[y=ae^{3x}+be^{-x}+\frac{42}{65}sin(2x)-\frac{11}5cos(2x)\]
...
as long as im not making any simple arithmatic erros, it does look to be headed in that direction

Answer 30

forgot to distribute the 1/13 to the -3cos(2x) ....

Answer 31

Yeah that does look better. You know where you're trying to find A'. Where does the 4 go?

Answer 32

... simple algebra mistakes is all, need to do a divide by 4, and a divide by 2 on B

that might help align this stuff betterly

Answer 33

A' = 1/4 e^(3x) cos(2x)

B' = -1/2 e^(x) cos(2x)

then integrate ....

Answer 34

ive got to focus on my paying job at the moment, ill run thru what ive posted later to see what other mistakes might have crept in

Answer 35

Aww That's so nice of you. Thank you once again.

Answer 36

yeah, i made a few basic errors while trying to type all that in the latex code :)

A' e^(3x) + B' e^(-x) = 0
A' 3e^(3x) - B' e^(-x) = cos(2x)

A' = 1/4 e^(-3x) cos(2x)
B' = -1/4 e^(x) cos(2x)

A = 1/52 e^(-3x) (2 sin(2x)-3 cos(2x))
B = -1/20 e^x (2 sin(2x)+cos(2x))

1/52 (2 sin(2x)-3 cos(2x)) -1/20 (2 sin(2x)+cos(2x))

which simplifies to:

-1/65 (7cos(2x)+4sin(2x))

Answer 37

Hey thanks! Wow you are a great helper!!

Answer 38

depending on your level of integrating skill, this might be a bit difficult integrating A' and B'

A more practical method is called "undetermined coefficients", and its pretty much a trial and error approach.

we have already determined the yh parts that goes to zero, so we would need to determine a suitable yp

since we want this to end up equaling cos(2x), make a guess at yp; yp = Acos(2x) + Bsin(2x), this time A and B are considered numbers, and not functions

Answer 39

Hey yeah I know the second method

Answer 40

Did some research and found out how to do it. However my mate told me that it doesn't always work. So I got quite stressed as an equation might come up where I cant use that method "/

Answer 41

yp = A cos(2x) + B sin(2x)
yp' = -2A sin(2x) + 2B cos(2x)
yp'' = -4A cos(2x) - 4B sin(2x)

plug them into our: y'' -2y' -3y = cos(2x)

-3yp = -3A cos(2x) -3B sin(2x)
-2yp' = 4A sin(2x) -4B cos(2x)
yp'' = -4A cos(2x) - 4B sin(2x)
------------------------------
cos(2x) = (-7A-4B) cos(2x) + (-7B+4A) sin(2x)

this only pans out if: 4A-7B = 0
and: -7A-4B = 1

this is again a system of equations in 2 unknowns, but no integration is needed, just algebra

Answer 42

come up in the exam*

Answer 43

true it doesnt always pan out, but it tends to be shorter and worth the effort to try :) there are tables printed of the most appropriate guesses for yp

Answer 44

Ohh okay. I'm scared now. I do get ur method BUT I find it really hard to integrate A' and B'

Answer 45

i found it a bit tedious to integrate it by hand and try to simplify it as well. Which is why i decided to suggest the undertermined coeff method:) most likely if you havent done the variation of parameters and wronskian, then you will only be asked diffyQs that relate to the best guess table

Answer 46

Yeah I haven't done that. I hope you're right. Fingers crossed. I feel better now. Thank you ^_^

Answer 47

good luck :)

Answer 48

Thank you :D