Question

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

Answer 1

I'd help you if I could, but when you solve this I posted one for you to help me with.

Answer 2

\[y=y_h+y_p\]
\[y_h=c_1 e^{2x}+c_2 e^{-2x}\]
its that yp thats making me loose my religion :)

Answer 3

Wooooooow. Just wondering, where did you get this problem from?

Answer 4

Where the hell did c come into play?

Answer 5

i call it #14 form section 4.3 ... from HELL!!! lol

Answer 6

@Calculus102 It's differential equations. It's like a class after Calc III

Answer 7

Calc 3??? Like multivariate calculus?

Answer 8

I'm done here, I am zero help.

Answer 9

I'm in PRE calculus. This is madness.

Answer 10

\[y_p=(ax^2+bx+c)sin(2x)+(fx^2+gx+h)cos(2x)\]

Answer 11

calciii is cake compared to having to deal with this tripe ....

Answer 12

Bernoulli's method, anyone?

Answer 13

this is very simple ,chain rule,product rule and some pellet mixed together

Answer 14

burnoodle aint gonna help on this one

Answer 15

wait nevermind... bernoulli was if u had y^n lol not nth derivative.. i suck at math

Answer 16

its keeping track of the mindnumbingness till the end ... i dont think i have a happy personality left in me after this lol

Answer 17

I took diff eq's.. use the char equation?
r^2-4r = 0
r(r-4)=0
r = 0, r = 4

Answer 18

but what about the right? i forgot.. and i only took this like a year ago.

Answer 19

yeah the homogenous parts a breeze done did that

Answer 20

the right is one of those "make a good guess" type deals

Answer 21

heres my guess
\[y_p=(ax^2+bx+c)sin(2x)+(fx^2+gx+h)cos(2x)\]

Answer 22

ohh one of the rules? anyway i gotta eat dinner, mom's yelling at me lol.

Answer 23

\[y_p=(ax^2+bx+c)sin(2x)+(fx^2+gx+h)cos(2x)\]
\[y' _p=(2ax+b)sin(2x)+(2ax^2+2bx+2c)cos(2x)\]\[\hspace{5em} +(2fx+g)cos(2x)+(-2fx^2-2gx-2h)sin(2x)\]
\[y'' _p=(2a)sin(2x)+(4ax+2b)cos(2x)\]\[\hspace{3em}+(4ax+2b)cos(2x)+(-4ax^2-4bx-4c)sin(2x)\]\[\hspace{6em} +(2f)cos(2x)+(-4fx-2g)sin(2x)\]\[\hspace{9em}+(-4fx-2g)sin(2x)+(-4fx^2-4gx-4h)cos(2x)\]

Answer 24

\[(-4ax^2-4bx-4c)sin(2x)\]\[(2a-4ax^2-4bx-4c-4fx-2g-4fx-2g)sin(2x)\]
\[(-4fx^2-4gx-4h)cos(2x)\]\[(4ax+2b+4qx+2b+2f-4fx^2-4gx-4h)cos(2x)\]

\[(-8ax^2-8bx-8c+2a-8fx-4)sin(2x)=(x^2-3)sin(2x)\]\[a=-\frac{1}{8}\]
\[(-8bx-8c-\frac{1}{4}-8fx-4)=(-3)\]
ugh ... messed it up again

\[(-8fx^2-8gx-8h+4b+8ax+4b+2f)cos(2x)=0cos(2x)\]

Answer 25

LOL...THIS CAN BE pelletY

Answer 26

satellite, where does your math knowledge come from?

Answer 27

Wait LOOOLL, u're amistre!!! Doesn't matter man, where does your knowledge come from amistre?

Answer 28

You guys are epic! Like seriously! I'm very impressed, look up to you guys :))!

Answer 29

\[(-8ax^2-8bx-8c+2a-8fx-4g)=(x^2-3)\]
\[(-8a)x^2+(-8b-8f)x+(2a-8c-4g)=x^2+0x-3\]
i gots no math knowledge ;)

Answer 30

well, I know a = -1/8 regardless of what that other stuff does :)

Answer 31

and b=-f ...

Answer 32

i spose we gotta undo the cosines now and have some systems of equations to play with for the remaing terms

Answer 33

(-8f) x^2+(-8g+8a)x+(-8h+8b+2f)=0x^2+0x+0

f = 0 so b = 0

Answer 34

which means h=0 too for that constant

Answer 35

(-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

Answer 36

2(-1/8)-8c-4(-1/8)=-3

-1/4 -8c +1/2=-3

-8c =-3' 1/4

c =13/32

Answer 37

\[y''-4y=(x^2-3)sin(2x)\]
\[y=c_1e^{2x}+c_2e^{-2x}-(\frac{1}{8}x^2-\frac{13}{32})sin(2x)-\frac{1}{8}xcos(2x)\]

Answer 38

and i believe the wolf agrees .... yay!!

Answer 39

if JamesJ had been here he could of answered it with one hand tied behind his back ;)

Answer 40

Amistre64? is that you?

Answer 41

remember me its gopeder :)

Answer 42

thats what it says on my nametag :)

Answer 43

Do you Remember me im the bear guy :)

Answer 44

i vaguely recall a screenname like yours; at my age its all i can do to remember to wear socks

Answer 45

xD we talked alot maybe 4 mounts ago

Answer 46

yeah, 4 months is an eternity around here lol how you been?

Answer 47

haha so true :) Not much Just starting Geometry.. its loads of fun

Answer 48

if you ever meet euclid, punch him in the breadbasket for me. hes to blame for alot of that geometry stuff

Answer 49

xD oh i will :P

Answer 50

But at the end of the day its really cool because it relates to real world

Answer 51

Amistre, sorry to bother you guys, but can you take a look at this? please?
http://openstudy.com/study#/updates/4f4d9f77e4b019d0ebade348

Answer 52

Refer to the attachment.

Answer 53

We got "derive" at the college that looks quite similar to that kind of thing.