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

- amistre64

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

- Stacey Warren - Expert brainly.com

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- chestercat

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- Lukecrayonz

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

- amistre64

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

- anonymous

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

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## More answers

- Lukecrayonz

Where the hell did c come into play?

- amistre64

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

- anonymous

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

- anonymous

Calc 3??? Like multivariate calculus?

- Lukecrayonz

I'm done here, I am zero help.

- Lukecrayonz

I'm in PRE calculus. This is madness.

- amistre64

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

- amistre64

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

- bahrom7893

Bernoulli's method, anyone?

- anonymous

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

- amistre64

burnoodle aint gonna help on this one

- bahrom7893

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

- amistre64

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

- bahrom7893

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

- bahrom7893

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

- amistre64

yeah the homogenous parts a breeze done did that

- amistre64

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

- amistre64

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

- bahrom7893

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

- amistre64

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

- amistre64

\[(-4ax^2-4bx-4c)sin(2x)\]\[(2a-4ax^2-4bx-4c-4fx-2g-4fx-2g)sin(2x)\]
\[(-4fx^2-4gx-4h)cos(2x)\]\[(4ax+2b+4qx+2b+2f-4fx^2-4gx-4h)cos(2x)\]
\[(-8ax^2-8bx-8c+2a-8fx-4)sin(2x)=(x^2-3)sin(2x)\]\[a=-\frac{1}{8}\]
\[(-8bx-8c-\frac{1}{4}-8fx-4)=(-3)\]
ugh ... messed it up again
\[(-8fx^2-8gx-8h+4b+8ax+4b+2f)cos(2x)=0cos(2x)\]

- anonymous

LOL...THIS CAN BE pelletY

- bahrom7893

satellite, where does your math knowledge come from?

- bahrom7893

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

- bahrom7893

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

- amistre64

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

- amistre64

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

- amistre64

and b=-f ...

- amistre64

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

- amistre64

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

- amistre64

which means h=0 too for that constant

- amistre64

(-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

- amistre64

2(-1/8)-8c-4(-1/8)=-3
-1/4 -8c +1/2=-3
-8c =-3' 1/4
c =13/32

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

Amistre64? is that you?

- anonymous

remember me its gopeder :)

- amistre64

thats what it says on my nametag :)

- anonymous

Do you Remember me im the bear guy :)

- amistre64

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

- anonymous

xD we talked alot maybe 4 mounts ago

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

xD oh i will :P

- anonymous

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

- bahrom7893

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

- anonymous

Refer to the attachment.

##### 1 Attachment

- amistre64

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

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