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\[y''-4y=(x^2-3)sin(2x)\] solve the diffyQ; oy what a nightmare ...

Mathematics
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I'd help you if I could, but when you solve this I posted one for you to help me with.
\[y=y_h+y_p\] \[y_h=c_1 e^{2x}+c_2 e^{-2x}\] its that yp thats making me loose my religion :)
Wooooooow. Just wondering, where did you get this problem from?

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Other answers:

Where the hell did c come into play?
i call it #14 form section 4.3 ... from HELL!!! lol
@Calculus102 It's differential equations. It's like a class after Calc III
Calc 3??? Like multivariate calculus?
I'm done here, I am zero help.
I'm in PRE calculus. This is madness.
\[y_p=(ax^2+bx+c)sin(2x)+(fx^2+gx+h)cos(2x)\]
calciii is cake compared to having to deal with this tripe ....
Bernoulli's method, anyone?
this is very simple ,chain rule,product rule and some pellet mixed together
burnoodle aint gonna help on this one
wait nevermind... bernoulli was if u had y^n lol not nth derivative.. i suck at math
its keeping track of the mindnumbingness till the end ... i dont think i have a happy personality left in me after this lol
I took diff eq's.. use the char equation? r^2-4r = 0 r(r-4)=0 r = 0, r = 4
but what about the right? i forgot.. and i only took this like a year ago.
yeah the homogenous parts a breeze done did that
the right is one of those "make a good guess" type deals
heres my guess \[y_p=(ax^2+bx+c)sin(2x)+(fx^2+gx+h)cos(2x)\]
ohh one of the rules? anyway i gotta eat dinner, mom's yelling at me lol.
\[y_p=(ax^2+bx+c)sin(2x)+(fx^2+gx+h)cos(2x)\] \[y' _p=(2ax+b)sin(2x)+(2ax^2+2bx+2c)cos(2x)\]\[\hspace{5em} +(2fx+g)cos(2x)+(-2fx^2-2gx-2h)sin(2x)\] \[y'' _p=(2a)sin(2x)+(4ax+2b)cos(2x)\]\[\hspace{3em}+(4ax+2b)cos(2x)+(-4ax^2-4bx-4c)sin(2x)\]\[\hspace{6em} +(2f)cos(2x)+(-4fx-2g)sin(2x)\]\[\hspace{9em}+(-4fx-2g)sin(2x)+(-4fx^2-4gx-4h)cos(2x)\]
\[(-4ax^2-4bx-4c)sin(2x)\]\[(2a-4ax^2-4bx-4c-4fx-2g-4fx-2g)sin(2x)\] \[(-4fx^2-4gx-4h)cos(2x)\]\[(4ax+2b+4qx+2b+2f-4fx^2-4gx-4h)cos(2x)\] \[(-8ax^2-8bx-8c+2a-8fx-4)sin(2x)=(x^2-3)sin(2x)\]\[a=-\frac{1}{8}\] \[(-8bx-8c-\frac{1}{4}-8fx-4)=(-3)\] ugh ... messed it up again \[(-8fx^2-8gx-8h+4b+8ax+4b+2f)cos(2x)=0cos(2x)\]
LOL...THIS CAN BE pelletY
satellite, where does your math knowledge come from?
Wait LOOOLL, u're amistre!!! Doesn't matter man, where does your knowledge come from amistre?
You guys are epic! Like seriously! I'm very impressed, look up to you guys :))!
\[(-8ax^2-8bx-8c+2a-8fx-4g)=(x^2-3)\] \[(-8a)x^2+(-8b-8f)x+(2a-8c-4g)=x^2+0x-3\] i gots no math knowledge ;)
well, I know a = -1/8 regardless of what that other stuff does :)
and b=-f ...
i spose we gotta undo the cosines now and have some systems of equations to play with for the remaing terms
(-8f) x^2+(-8g+8a)x+(-8h+8b+2f)=0x^2+0x+0 f = 0 so b = 0
which means h=0 too for that constant
(-8(0)) x^2+(-8g+8a)x+(-8(0)+8(0)+2(0))=0x^2+0x+0 -8g+8a = 0; and a = -1/8 -8g-1 = 0 ; g = -1/8
2(-1/8)-8c-4(-1/8)=-3 -1/4 -8c +1/2=-3 -8c =-3' 1/4 c =13/32
\[y''-4y=(x^2-3)sin(2x)\] \[y=c_1e^{2x}+c_2e^{-2x}-(\frac{1}{8}x^2-\frac{13}{32})sin(2x)-\frac{1}{8}xcos(2x)\]
and i believe the wolf agrees .... yay!!
if JamesJ had been here he could of answered it with one hand tied behind his back ;)
Amistre64? is that you?
remember me its gopeder :)
thats what it says on my nametag :)
Do you Remember me im the bear guy :)
i vaguely recall a screenname like yours; at my age its all i can do to remember to wear socks
xD we talked alot maybe 4 mounts ago
yeah, 4 months is an eternity around here lol how you been?
haha so true :) Not much Just starting Geometry.. its loads of fun
if you ever meet euclid, punch him in the breadbasket for me. hes to blame for alot of that geometry stuff
xD oh i will :P
But at the end of the day its really cool because it relates to real world
Amistre, sorry to bother you guys, but can you take a look at this? please? http://openstudy.com/study#/updates/4f4d9f77e4b019d0ebade348
Refer to the attachment.
1 Attachment
We got "derive" at the college that looks quite similar to that kind of thing.

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