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

- anonymous

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- chestercat

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- anonymous

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

- anonymous

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

- anonymous

@satellite73

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## More answers

- anonymous

really i am sorry i have no clue

- anonymous

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

- anonymous

So what should I do?

- anonymous

Did you try this using wronskian

- anonymous

I don't know what that is

- anonymous

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

- anonymous

did you look up wronskian

- across

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.

- anonymous

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

- anonymous

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

- across

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

- across

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

- TuringTest

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

- TuringTest

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

- TuringTest

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

- TuringTest

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

- across

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).

- TuringTest

@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)

- anonymous

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

- TuringTest

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

- TuringTest

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

- TuringTest

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

- across

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

- TuringTest

thanks!
Gotta get sharp on that technique

- anonymous

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

- TuringTest

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

- anonymous

correct

- TuringTest

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

- anonymous

Asin2x+Bcos2x?

- TuringTest

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

- anonymous

ok

- TuringTest

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

- TuringTest

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

- TuringTest

linearly independent*

- TuringTest

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)

- anonymous

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

- across

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

- TuringTest

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

- anonymous

A=1

- TuringTest

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

- anonymous

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

- anonymous

Thanks to you too @across

- anonymous

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

- anonymous

?

- TuringTest

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.

- TuringTest

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

- anonymous

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

- anonymous

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

- TuringTest

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

- anonymous

No hurry =P

- TuringTest

I did get A=-1/2 by hand

- TuringTest

actually you can see that fairly quickly...

- TuringTest

wanna see how?

- anonymous

sure

- anonymous

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

- TuringTest

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 :/

- TuringTest

okay I have confirmed wolf's answer on paper

- anonymous

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

- anonymous

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

- TuringTest

you wanna walk through it?

- anonymous

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

- TuringTest

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

- anonymous

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

- TuringTest

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

- anonymous

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

- TuringTest

yep

- anonymous

something is still wrong though

- TuringTest

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

- anonymous

oh I see what I did

- TuringTest

sweet

- anonymous

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

- TuringTest

no it aint :P

- TuringTest

you changed a sign

- TuringTest

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

- anonymous

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

- TuringTest

yay!

- anonymous

LATEX makes all the difference!

- TuringTest

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

- TuringTest

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

- TuringTest

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

- TuringTest

I mean in the complimentary

- across

Hopefully this summarizes everything that went on in here nicely for you :)
https://dl.dropbox.com/u/12189012/OS.pdf

- anonymous

oh my gosh! did you just make that?

- across

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

- TuringTest

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

- anonymous

you are soo sweet! thank you!

- anonymous

OS rocks!

- TuringTest

yes it does :)

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