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

- amistre64

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

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.