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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now on to yp_2, where I'm having some trouble...here is how far I've gotten.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0really i am sorry i have no clue

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there 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
 3 years ago
Best ResponseYou've already chosen the best response.0Did you try this using wronskian

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't know what that is

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can 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
 3 years ago
Best ResponseYou've already chosen the best response.0did you look up wronskian

across
 3 years ago
Best ResponseYou've already chosen the best response.1You 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.

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

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

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

across
 3 years ago
Best ResponseYou've already chosen the best response.1In your case, \(\alpha=0\) and \(\beta=2\).

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2I'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
 3 years ago
Best ResponseYou've already chosen the best response.2the guess for \(\cos(Ax)\) or \(\sin(Ax)\) is \(y_p=B\cos(Ax)+C\sin(Ax)\)

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2not sure why you picked up the extra x on \(y _2\)

across
 3 years ago
Best ResponseYou've already chosen the best response.1That'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
 3 years ago
Best 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)

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2or look quickly at example 8d here (not that POMN is that thorough...)

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2as was said, it's the superposition principle at work is all

across
 3 years ago
Best ResponseYou've already chosen the best response.1That'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
 3 years ago
Best ResponseYou've already chosen the best response.2thanks! Gotta get sharp on that technique

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2and if it were just g(x)=sin(2x) what would the guess be?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2right, 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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2so 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
 3 years ago
Best ResponseYou've already chosen the best response.2since 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
 3 years ago
Best ResponseYou've already chosen the best response.2linearly independent*

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2hence 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
 3 years ago
Best ResponseYou've already chosen the best response.0I understand it now! Is that it?....or I still gotta solve for A B and C of course

across
 3 years ago
Best ResponseYou've already chosen the best response.1No! That's totally fine. :) I couldn't have said it better.

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0LOL....Yeah I got this from here on. Thanks Turing!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks to you too @across

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so I can skip the annihilator method (I rather avoid it because it looks confusing)

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2it 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
 3 years ago
Best ResponseYou've already chosen the best response.2http://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
 3 years ago
Best ResponseYou've already chosen the best response.0Hhmmmm....I'll do it manually too and compare

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

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2I did get A=1/2 by hand

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2actually you can see that fairly quickly...

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2you 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
 3 years ago
Best ResponseYou've already chosen the best response.2okay I have confirmed wolf's answer on paper

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I've gotten A=1/2 B=1/4 C=5/2 and D=5/2

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2you wanna walk through it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now 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
 3 years ago
Best ResponseYou've already chosen the best response.2nah, still got problems with C and D can you write out yp, yp', and yp'' ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y_p=Ax+B+Csin2x+Dcos2x\] \[y_p'=A+2Ccos2x2Dsin2x\] \[y_p''=4Csin2x4Dcos2x\]

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2thanks 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[(4C2D2C)=1\] \[(4D+2C2D)=0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0something is still wrong though

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2not yet, you just must have messed up solving the system

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2the first equation is 2D, not +2D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that'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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0LATEX makes all the difference!

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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2the 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
 3 years ago
Best ResponseYou've already chosen the best response.2I mean in the complimentary

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh my gosh! did you just make that?

across
 3 years ago
Best ResponseYou've already chosen the best response.1I felt like practicing my rusty \(\LaTeX\) skills.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you are soo sweet! thank you!
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