Here's the question you clicked on:
MathSofiya
\[y''+y'-2y=x+sin2x\] \[y(0)=1\] \[y'(0)=0\] auxiliary equation: \[r^2+r-2=0\] \[r_1=-2, r_2=1\] \[y_c(x)=C_1e^{-2x}+C_2e^x\] \[y''+y'-2y=x\] \[y_{p1}(x)=Ax+B\] \[y_{p1}'(x)=A\] \[y_{p1}''(x)=0\] \[A=-\frac12\] \[B=-1\] \[y_{p1}=-\frac12x-1\]
Now on to yp_2, where I'm having some trouble...here is how far I've gotten.
\[y''+y'-2y=sin2x\] \[y_{p2}(x)=Axcos2x+Bxsin2x\] \[y_{p2}'(x)=2Acos2x-2Axsin2x+2Bsin2x+2Bxcos2x\] \[y_{p2}''(x)=-4Asin2x-4Axcos2x+4Bcos2x-4Bxsin2x\]
really i am sorry i have no clue
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
So what should I do?
Did you try this using wronskian
I don't know what that is
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
did you look up wronskian
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.
why is it \[D^2+4\]? Is it standard for the annihilator method or where did the 4 come from?
I''m watching Khan's academy videos right now...I should have a better understanding in an hour or so
The differential operator\[\left[D^2-2\alpha D+\left(\alpha^2+\beta^2\right)\right]^n\]anihilates each of the functions\[e^{\alpha x}\cos\beta x, xe^{\alpha x}\cos\beta x, \dots, x^{n-1}e^{\alpha x}\cos\beta x,\\e^{\alpha x}\sin\beta x, xe^{\alpha x}\sin\beta x, \dots, x^{n-1}e^{\alpha x}\sin\beta x.\]
In your case, \(\alpha=0\) and \(\beta=2\).
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
the guess for \(\cos(Ax)\) or \(\sin(Ax)\) is \(y_p=B\cos(Ax)+C\sin(Ax)\)
so add your two together and get\[y_p=Ax+B+C\cos(2x)+D\cos(2x)\]
not sure why you picked up the extra x on \(y _2\)
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).
@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)
I still have two more videos to watch...brb, (hopefully it will make more sense then...)
or look quickly at example 8-d here (not that POMN is that thorough...)
http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
as was said, it's the superposition principle at work is all
That's right, @TuringTest. Given \(\sin2x\), applying the differential operator \(D^2+4\) to it yields:\[\begin{align}(D^2+4)\sin2x&=D^2\sin2x+4\sin2x\\&=2D\cos2x+4\sin2x\\&=-4\sin2x+4\sin2x\\&=0.\end{align}\]:)
thanks! Gotta get sharp on that technique
Let's start from \[y''+y'-2y=x+sin2x\] and \[y_p\] It's still not making much sense
if it were just g(x)=x our guess would be Yp=Ax+B, right?
and if it were just g(x)=sin(2x) what would the guess be?
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
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
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
linearly independent*
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)
I understand it now! Is that it?....or I still gotta solve for A B and C of course
No! That's totally fine. :) I couldn't have said it better.
thanks :) so yeah, now it's business as usual (certainly a bit tedious, but straightforward I think)
I honestly don't feel like checking your work apart from wolfram, which is not always reliable, but hopefully you will agree.
LOL....Yeah I got this from here on. Thanks Turing!
Thanks to you too @across
so I can skip the annihilator method (I rather avoid it because it looks confusing)
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.
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
Hhmmmm....I'll do it manually too and compare
I worked on understanding this idea for 2 days now...thanks guys for helping me finally understand it, honestly!
That's the best reward I could hope for, happy to help. I'll let you know what I get manually... eventually ;)
I did get A=-1/2 by hand
actually you can see that fairly quickly...
I forgot to take the first and second derivative before solving for A B and C...that's why I said A=1 OOOooops
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 :/
okay I have confirmed wolf's answer on paper
I've gotten A=-1/2 B=-1/4 C=5/2 and D=-5/2
I guess my C's and D's are wrong...oh well I'll take a closer look at wolfram
you wanna walk through it?
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
nah, still got problems with C and D can you write out yp, yp', and yp'' ?
\[y_p=Ax+B+Csin2x+Dcos2x\] \[y_p'=A+2Ccos2x-2Dsin2x\] \[y_p''=-4Csin2x-4Dcos2x\]
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
\[(-4C-2D-2C)=1\] \[(-4D+2C-2D)=0\]
something is still wrong though
not yet, you just must have messed up solving the system
nope not yet C=3D and then you sub it into the first equation -6(3D)+2D=1 that still is D=-1/16
the first equation is -2D, not +2D
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
LATEX makes all the difference!
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
btw I made the \(exact\) same mistake and got D=1/16 first
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
I mean in the complimentary
Hopefully this summarizes everything that went on in here nicely for you :) https://dl.dropbox.com/u/12189012/OS.pdf
oh my gosh! did you just make that?
I felt like practicing my rusty \(\LaTeX\) skills.
That is quite generous, and looks much nicer than my two pages of scribbles with things crossed out ;)
you are soo sweet! thank you!