Question

\[y''+y'-2y=x+sin2x\]
\[y(0)=1\]
\[y'(0)=0\]
auxiliary equation:
\[r^2+r-2=0\]
\[r_1=-2, r_2=1\]
\[y_c(x)=C_1e^{-2x}+C_2e^x\]
\[y''+y'-2y=x\]
\[y_{p1}(x)=Ax+B\]
\[y_{p1}'(x)=A\]
\[y_{p1}''(x)=0\]
\[A=-\frac12\]
\[B=-1\]
\[y_{p1}=-\frac12x-1\]

Answer 1

Now on to yp_2, where I'm having some trouble...here is how far I've gotten.

Answer 2

\[y''+y'-2y=sin2x\]
\[y_{p2}(x)=Axcos2x+Bxsin2x\]
\[y_{p2}'(x)=2Acos2x-2Axsin2x+2Bsin2x+2Bxcos2x\]
\[y_{p2}''(x)=-4Asin2x-4Axcos2x+4Bcos2x-4Bxsin2x\]

Answer 3

@satellite73

Answer 4

really i am sorry i have no clue

Answer 5

there is no need to select Ax cos 2x + Bx sin 2x the reason is , if you take out x as a common factor, then that "x" is actually part of a solution so you are bound to get incorrect solutions

Answer 6

So what should I do?

Answer 7

Did you try this using wronskian

Answer 8

I don't know what that is

Answer 9

can you use wikipedia to understand that method, use the keywords "wronskian equation" or "wronskian determinant"

I would use that approach to solve, I will use the method followed by you only under the cases where wronskian is 0

Answer 10

did you look up wronskian

Answer 11

You made a mistake assuming the form of \(y_{p_2}\):

I don't know if you've heard of the concept of an "annihilator" before, but the differential operator \((D^2+4)\) annihilates \(\sin2x\).

If we apply it to your ODE, we get:\[\begin{align}(D^2+4)(D^2+D-2)y&=(D^2+4)\sin2x,\\(m^2+4)(m^2+m-2)&=0,\\(m\pm2i)(m+2)(m-1)&=0.\end{align}\]Hence, \(y_{p_2}=A\cos2x+B\sin2x\).

This is different from what you got up there; notice that there is no \(x\) multiplying neither \(\cos\) nor \(\sin\). Now try and try and wrap it up as you would from here.

Answer 12

why is it \[D^2+4\]? Is it standard for the annihilator method or where did the 4 come from?

Answer 13

I''m watching Khan's academy videos right now...I should have a better understanding in an hour or so

Answer 14

The differential operator\[\left[D^2-2\alpha D+\left(\alpha^2+\beta^2\right)\right]^n\]anihilates each of the functions\[e^{\alpha x}\cos\beta x, xe^{\alpha x}\cos\beta x, \dots, x^{n-1}e^{\alpha x}\cos\beta x,\\e^{\alpha x}\sin\beta x, xe^{\alpha x}\sin\beta x, \dots, x^{n-1}e^{\alpha x}\sin\beta x.\]

Answer 15

In your case, \(\alpha=0\) and \(\beta=2\).

Answer 16

I'm sure @across knows exactly what she's talking about and I don't, but I would just make a single \(y_p\) which is the sum of the two individual ones

Answer 17

the guess for \(\cos(Ax)\) or \(\sin(Ax)\) is \(y_p=B\cos(Ax)+C\sin(Ax)\)

Answer 18

so add your two together and get\[y_p=Ax+B+C\cos(2x)+D\cos(2x)\]

Answer 19

not sure why you picked up the extra x on \(y
_2\)

Answer 20

That's correct. It follows from the principle of superposition given differential operators are linear transformations. :) I was thrown off a little bit when I saw @MathSofiya formulating two particular solutions, but then told myself: "heck, it's the same thing anyway." So I went with her approach, but I'd also combine them.

@TuringTest, that's right, but the term with the \(D\) coefficient should be a \(\sin\) (you could invert them, too, it doesn't matter).

Answer 21

@across oh yes, of of course, typo :P
I'm also now remembering the annihilator thing you did, the need to add the 4 coming from differntiating sin(2x) twice and getting -4 right?
(sorry to interject, just trying to remember)

Answer 22

I still have two more videos to watch...brb, (hopefully it will make more sense then...)

Answer 23

or look quickly at example 8-d here (not that POMN is that thorough...)

Answer 24

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

Answer 25

as was said, it's the superposition principle at work is all

Answer 26

That's right, @TuringTest.

Given \(\sin2x\), applying the differential operator \(D^2+4\) to it yields:\[\begin{align}(D^2+4)\sin2x&=D^2\sin2x+4\sin2x\\&=2D\cos2x+4\sin2x\\&=-4\sin2x+4\sin2x\\&=0.\end{align}\]:)

Answer 27

thanks!
Gotta get sharp on that technique

Answer 28

Let's start from
\[y''+y'-2y=x+sin2x\] and \[y_p\]
It's still not making much sense

Answer 29

if it were just g(x)=x our guess would be Yp=Ax+B, right?

Answer 30

correct

Answer 31

and if it were just g(x)=sin(2x) what would the guess be?

Answer 32

Asin2x+Bcos2x?

Answer 33

right, but let's call A=C and B=D so as not to confuse them with the other guess; they are not the same constants

Answer 34

ok

Answer 35

so for g(x)=x+sin(2x)
Yp=(Ax+B)+(Csin(2x)+Dsin(2x))
which is just the sum (superposition/linear combination) of our two guesses

Answer 36

since nothing on the right side is a linear combination f anything in the complimentary solution, your solutions are all linearly dependent
i.e. we don't need to screw with them any more by adding more x's or anything

Answer 37

linearly independent*

Answer 38

hence the guess for the particular is\[y_p=Ax+B+C\sin(2x)+B\cos(2x)\]the guess for the g(x)=x part+ the guess for the g(x)=sin(2x) part
making a little more sense
(sorry that I seemed to have discouraged @across from giving her I'm sure much more insightful answer)

Answer 39

I understand it now! Is that it?....or I still gotta solve for A B and C of course

Answer 40

No! That's totally fine. :) I couldn't have said it better.

Answer 41

thanks :)
so yeah, now it's business as usual (certainly a bit tedious, but straightforward I think)

Answer 42

A=1

Answer 43

I honestly don't feel like checking your work apart from wolfram, which is not always reliable, but hopefully you will agree.

Answer 44

LOL....Yeah I got this from here on. Thanks Turing!

Answer 45

Thanks to you too @across

Answer 46

so I can skip the annihilator method (I rather avoid it because it looks confusing)

Answer 47

?

Answer 48

it looks like wolfram wants to say A=-1/2
I will play with the system to find the constants, but I'm not going to rush it.

Answer 49

http://www.wolframalpha.com/input/?i=y''%2By'-2y%3Dx%2Bsin(2x)%2C%20y(0)%3D1%2C%20y'(0)%3D0&t=crmtb01
though I really don't trust wolf that much on DE's

Answer 50

Hhmmmm....I'll do it manually too and compare

Answer 51

I worked on understanding this idea for 2 days now...thanks guys for helping me finally understand it, honestly!

Answer 52

That's the best reward I could hope for, happy to help.
I'll let you know what I get manually... eventually ;)

Answer 53

No hurry =P

Answer 54

I did get A=-1/2 by hand

Answer 55

actually you can see that fairly quickly...

Answer 56

wanna see how?

Answer 57

sure

Answer 58

I forgot to take the first and second derivative before solving for A B and C...that's why I said A=1 OOOooops

Answer 59

you can also notice that the only part that will have an x left is Yp
and since you have -2Yp you will have -2Ax on the left and x on the right
hence A=-1/2
the rest is getting a bit frustrating :/

Answer 60

okay I have confirmed wolf's answer on paper

Answer 61

I've gotten A=-1/2 B=-1/4 C=5/2 and D=-5/2

Answer 62

I guess my C's and D's are wrong...oh well I'll take a closer look at wolfram

Answer 63

you wanna walk through it?

Answer 64

Now I got A=-1/2 B=-1/4 C=-1/4 D=1/4 that's before considering y(0)=1 and y'(0)=0

Answer 65

nah, still got problems with C and D
can you write out yp, yp', and yp'' ?

Answer 66

\[y_p=Ax+B+Csin2x+Dcos2x\]
\[y_p'=A+2Ccos2x-2Dsin2x\]
\[y_p''=-4Csin2x-4Dcos2x\]

Answer 67

thanks now just write out the result of doing the sines and cosines in y''+y'-2y
ignore the Ax an B as you have already solved it and it won't affect the C and D

Answer 68

\[(-4C-2D-2C)=1\]
\[(-4D+2C-2D)=0\]

Answer 69

yep

Answer 70

something is still wrong though

Answer 71

not yet, you just must have messed up solving the system

Answer 72

oh I see what I did

Answer 73

sweet

Answer 74

nope not yet
C=3D and then you sub it into the first equation
-6(3D)+2D=1 that still is D=-1/16

Answer 75

no it aint :P

Answer 76

you changed a sign

Answer 77

the first equation is -2D, not +2D

Answer 78

that's what I had right?
\[(-4C-2D-2C)=1\]
\[(-4D+2C-2D)=0\]
simplified
\[(-6C-2D)=1\]
\[(-6D+2C)=0\]
multiply the lower equation by 3
\[(-6C-2D)=1\]
\[(+6C-18D)=0\]
-----------------------+
-20D=1
D=-1/20

Answer 79

yay!

Answer 80

LATEX makes all the difference!

Answer 81

I really think that if programs are meant to do anything tedious in math, systems are about the most perfect example there is
although, a few linear algebra tricks can solve these 2x2's pretty darn fast

Answer 82

btw I made the \(exact\) same mistake and got D=1/16 first

Answer 83

the system to find the constants in the particular is actually a bit worse I'd say, so I used some linear algebra stuff there

Answer 84

I mean in the complimentary

Answer 85

Hopefully this summarizes everything that went on in here nicely for you :)

https://dl.dropbox.com/u/12189012/OS.pdf

Answer 86

oh my gosh! did you just make that?

Answer 87

I felt like practicing my rusty \(\LaTeX\) skills.

Answer 88

That is quite generous, and looks much nicer than my two pages of scribbles with things crossed out ;)

Answer 89

you are soo sweet! thank you!

Answer 90

OS rocks!

Answer 91

yes it does :)