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

really i am sorry i have no clue

So what should I do?

Did you try this using wronskian

I don't know what that is

did you look up wronskian

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

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

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 \(y
_2\)

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

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?

correct

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

Asin2x+Bcos2x?

ok

linearly independent*

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)

A=1

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

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

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

No hurry =P

I did get A=-1/2 by hand

actually you can see that fairly quickly...

wanna see how?

sure

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'' ?

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

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

yep

something is still wrong though

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

oh I see what I did

sweet

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

no it aint :P

you changed a sign

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

yay!

LATEX makes all the difference!

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

I mean in the complimentary

oh my gosh! did you just make that?

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

you are soo sweet! thank you!

OS rocks!

yes it does :)