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\[y''+y'2y=x+sin2x\]
\[y(0)=1\]
\[y'(0)=0\]
auxiliary equation:
\[r^2+r2=0\]
\[r_1=2, r_2=1\]
\[y_c(x)=C_1e^{2x}+C_2e^x\]
\[y''+y'2y=x\]
\[y_{p1}(x)=Ax+B\]
\[y_{p1}'(x)=A\]
\[y_{p1}''(x)=0\]
\[A=\frac12\]
\[B=1\]
\[y_{p1}=\frac12x1\]
 one year ago
 one year ago
\[y''+y'2y=x+sin2x\] \[y(0)=1\] \[y'(0)=0\] auxiliary equation: \[r^2+r2=0\] \[r_1=2, r_2=1\] \[y_c(x)=C_1e^{2x}+C_2e^x\] \[y''+y'2y=x\] \[y_{p1}(x)=Ax+B\] \[y_{p1}'(x)=A\] \[y_{p1}''(x)=0\] \[A=\frac12\] \[B=1\] \[y_{p1}=\frac12x1\]
 one year ago
 one year ago

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MathSofiyaBest ResponseYou've already chosen the best response.0
Now on to yp_2, where I'm having some trouble...here is how far I've gotten.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
\[y''+y'2y=sin2x\] \[y_{p2}(x)=Axcos2x+Bxsin2x\] \[y_{p2}'(x)=2Acos2x2Axsin2x+2Bsin2x+2Bxcos2x\] \[y_{p2}''(x)=4Asin2x4Axcos2x+4Bcos2x4Bxsin2x\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
really i am sorry i have no clue
 one year ago

psi9epsilonBest ResponseYou've already chosen the best response.0
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
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
So what should I do?
 one year ago

psi9epsilonBest ResponseYou've already chosen the best response.0
Did you try this using wronskian
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
I don't know what that is
 one year ago

psi9epsilonBest ResponseYou've already chosen the best response.0
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
 one year ago

psi9epsilonBest ResponseYou've already chosen the best response.0
did you look up wronskian
 one year ago

acrossBest ResponseYou've already chosen the best response.1
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:\[\begin{align}(D^2+4)(D^2+D2)y&=(D^2+4)\sin2x,\\(m^2+4)(m^2+m2)&=0,\\(m\pm2i)(m+2)(m1)&=0.\end{align}\]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.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
why is it \[D^2+4\]? Is it standard for the annihilator method or where did the 4 come from?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
I''m watching Khan's academy videos right now...I should have a better understanding in an hour or so
 one year ago

acrossBest ResponseYou've already chosen the best response.1
The differential operator\[\left[D^22\alpha D+\left(\alpha^2+\beta^2\right)\right]^n\]anihilates each of the functions\[e^{\alpha x}\cos\beta x, xe^{\alpha x}\cos\beta x, \dots, x^{n1}e^{\alpha x}\cos\beta x,\\e^{\alpha x}\sin\beta x, xe^{\alpha x}\sin\beta x, \dots, x^{n1}e^{\alpha x}\sin\beta x.\]
 one year ago

acrossBest ResponseYou've already chosen the best response.1
In your case, \(\alpha=0\) and \(\beta=2\).
 one year ago

TuringTestBest ResponseYou've already chosen the best response.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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
the guess for \(\cos(Ax)\) or \(\sin(Ax)\) is \(y_p=B\cos(Ax)+C\sin(Ax)\)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
so add your two together and get\[y_p=Ax+B+C\cos(2x)+D\cos(2x)\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
not sure why you picked up the extra x on \(y _2\)
 one year ago

acrossBest ResponseYou've already chosen the best response.1
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).
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
@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)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
I still have two more videos to watch...brb, (hopefully it will make more sense then...)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
or look quickly at example 8d here (not that POMN is that thorough...)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
as was said, it's the superposition principle at work is all
 one year ago

acrossBest ResponseYou've already chosen the best response.1
That's right, @TuringTest. Given \(\sin2x\), applying the differential operator \(D^2+4\) to it yields:\[\begin{align}(D^2+4)\sin2x&=D^2\sin2x+4\sin2x\\&=2D\cos2x+4\sin2x\\&=4\sin2x+4\sin2x\\&=0.\end{align}\]:)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
thanks! Gotta get sharp on that technique
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
Let's start from \[y''+y'2y=x+sin2x\] and \[y_p\] It's still not making much sense
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
if it were just g(x)=x our guess would be Yp=Ax+B, right?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
and if it were just g(x)=sin(2x) what would the guess be?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
linearly independent*
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
I understand it now! Is that it?....or I still gotta solve for A B and C of course
 one year ago

acrossBest ResponseYou've already chosen the best response.1
No! That's totally fine. :) I couldn't have said it better.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
thanks :) so yeah, now it's business as usual (certainly a bit tedious, but straightforward I think)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
I honestly don't feel like checking your work apart from wolfram, which is not always reliable, but hopefully you will agree.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
LOL....Yeah I got this from here on. Thanks Turing!
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
Thanks to you too @across
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
so I can skip the annihilator method (I rather avoid it because it looks confusing)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
Hhmmmm....I'll do it manually too and compare
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
I worked on understanding this idea for 2 days now...thanks guys for helping me finally understand it, honestly!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
That's the best reward I could hope for, happy to help. I'll let you know what I get manually... eventually ;)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
I did get A=1/2 by hand
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
actually you can see that fairly quickly...
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
I forgot to take the first and second derivative before solving for A B and C...that's why I said A=1 OOOooops
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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 :/
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
okay I have confirmed wolf's answer on paper
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
I've gotten A=1/2 B=1/4 C=5/2 and D=5/2
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
I guess my C's and D's are wrong...oh well I'll take a closer look at wolfram
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
you wanna walk through it?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
nah, still got problems with C and D can you write out yp, yp', and yp'' ?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
\[y_p=Ax+B+Csin2x+Dcos2x\] \[y_p'=A+2Ccos2x2Dsin2x\] \[y_p''=4Csin2x4Dcos2x\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
\[(4C2D2C)=1\] \[(4D+2C2D)=0\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
something is still wrong though
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
not yet, you just must have messed up solving the system
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
oh I see what I did
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
nope not yet C=3D and then you sub it into the first equation 6(3D)+2D=1 that still is D=1/16
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
the first equation is 2D, not +2D
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
that's what I had right? \[(4C2D2C)=1\] \[(4D+2C2D)=0\] simplified \[(6C2D)=1\] \[(6D+2C)=0\] multiply the lower equation by 3 \[(6C2D)=1\] \[(+6C18D)=0\] + 20D=1 D=1/20
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
LATEX makes all the difference!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
btw I made the \(exact\) same mistake and got D=1/16 first
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
I mean in the complimentary
 one year ago

acrossBest ResponseYou've already chosen the best response.1
Hopefully this summarizes everything that went on in here nicely for you :) https://dl.dropbox.com/u/12189012/OS.pdf
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
oh my gosh! did you just make that?
 one year ago

acrossBest ResponseYou've already chosen the best response.1
I felt like practicing my rusty \(\LaTeX\) skills.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
That is quite generous, and looks much nicer than my two pages of scribbles with things crossed out ;)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
you are soo sweet! thank you!
 one year ago
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