so I have
y'' = e^x((1+√2) cos√2x + (1-√2)sin√2x) + e^x(2-√2)sin√2x + (√2 -2)cos√2x)

y' = -2e^x((1+√2)cos√2x + (1-√2)sin√2 x)

y = 3e^x(cos√2x + sin √2x)

and i have to prove that y''=2y'-3y

and I'm having just a little bit of trouble =/
also writing this out was a pain in the ass on a comp

Fr33t

United States1128 Posts

March 13 2009 04:36 GMT

Were those three equations given to you or did you derive/integrate them yourself?

Yogurt

United States4258 Posts

March 13 2009 04:38 GMT

On March 13 2009 13:36 Fr33t wrote:
Were those three equations given to you or did you derive/integrate them yourself?

the teacher ended up deriving them for us

Fr33t

United States1128 Posts

March 13 2009 04:40 GMT

So I assume you are given just y right?

Yogurt

United States4258 Posts

March 13 2009 04:42 GMT

yup its up there in my post

Fr33t

United States1128 Posts

March 13 2009 04:44 GMT

I don't know, I'm deriving it right now myself because y' and y" just don't seem right. Or I could be completely wrong, it's late.

Cloud

Sexico5880 Posts

March 13 2009 04:46 GMT

Well shit use an equation editor and post a screenshot of it.

micronesia

United States24725 Posts

March 13 2009 04:55 GMT

Er that's just plug and chug... not a proof. You can plug it into an algebraic math program to verify it, but doing it by hand should just be tedious.

Fr33t

United States1128 Posts

March 13 2009 04:58 GMT

No one else is trying it?

Yogurt

United States4258 Posts

March 13 2009 05:03 GMT

#10

On March 13 2009 13:55 micronesia wrote:
Er that's just plug and chug... not a proof. You can plug it into an algebraic math program to verify it, but doing it by hand should just be tedious.

regardless, my plug and chug skills must be terrible 'cus i can't figure it out for the life of me, if you wouldn't mind helping me a long a little bit i'd be extremely grateful

i'm trying to get it into an equation editor as well but the square roots are giving me trouble

Fr33t

United States1128 Posts

March 13 2009 05:09 GMT

#11

+ Show Spoiler [XMaxima] +

(%i6) integrate(3*e^(x*(cos(sqrt(2*x))+sin(sqrt(2*x)))),x);
(%o6)
/
[ (sin(sqrt(2) sqrt(x)) + cos(sqrt(2) sqrt(x))) x
3 I e dx
]
/

Interesting...

Yogurt

United States4258 Posts

March 13 2009 05:20 GMT

#12

On March 13 2009 14:09 Fr33t wrote:
+ Show Spoiler [XMaxima] +

(%i6) integrate(3*e^(x*(cos(sqrt(2*x))+sin(sqrt(2*x)))),x);
(%o6)
/
[ (sin(sqrt(2) sqrt(x)) + cos(sqrt(2) sqrt(x))) x
3 I e dx
]
/

Interesting...

I don't get it =?

also uploaded scanned picture

Fr33t

United States1128 Posts

March 13 2009 05:24 GMT

#13

oh...I thought you meant √(2x) not √2 * x

Yogurt

United States4258 Posts

March 13 2009 05:25 GMT

#14

On March 13 2009 14:24 Fr33t wrote:
oh...I thought you meant √(2x) not √2 * x

i think its the latter, but i'm not entirely a 100% on that =/

Fr33t

United States1128 Posts

March 13 2009 05:26 GMT

#15

Now I get this

(%i9) integrate(3*e^(x*(cos(sqrt(2))*x+sin(sqrt(2))*x)),x);
(%o9)
3 sqrt(%pi) erf(sqrt(sin(sqrt(2)) + cos(sqrt(2))) sqrt(- log(e)) x)
-------------------------------------------------------------------
2 sqrt(sin(sqrt(2)) + cos(sqrt(2))) sqrt(- log(e))

Yogurt

United States4258 Posts

March 13 2009 05:40 GMT

#16

i'm just gonna go to sleep, thanks anyway fr33t

Fr33t

United States1128 Posts

March 13 2009 05:45 GMT

#17

Anytime, that problem is ridiculous. I took AB calc last year and we never got something like that.

Malongo

Chile3472 Posts

March 13 2009 05:50 GMT

#18

y = 3e^x(cos√2x + sin √2x) = 3e^x*2/√2*(√2/2*cos√2x + √2/2*sin √2x)=
3e^x*2/√2*sin(√2x+Pi/4) Now derivate 2 times and find
y'= 3e^x*2/√2*sin(√2x+Pi/4)+ 3e^x*2*cos(√2x+Pi/4)
y''= 3e^x*2/√2*sin(√2x+Pi/4)+ 3e^x*2*cos(√2x+Pi/4) + 3e^x*2*cos(√2x+Pi/4) - 3e^x*2*√2sin(√2x+Pi/4)=2*3e^x*2*cos(√2x+Pi/4)-3e^x*√2sin(√2x+Pi/4)

2y'-3y= 6e^x*2*cos(√2x+Pi/4)-3e^x*2/√2*sin(√2x+Pi/4) = y''

easy! READ THE FUCKING second LINE

Malongo

Chile3472 Posts

March 13 2009 05:52 GMT

#19

And you owe me 20 bucks or something...

Fr33t

United States1128 Posts

March 13 2009 06:04 GMT

#20

^.-

Wow...that's impossible to read. Still, you did it different than his teacher since I do see pi in your work but not in the OP. I'm done.

Turbovolver

Australia2394 Posts

March 13 2009 06:08 GMT

#21

LOL, the original problem was written so unclearly, because as far as I can tell the 3 and the -2 have already been pre-multiplied. You should just be adding the three things together, not subbing them in to y'' = 2y' - 3y.

But hell, that ODE is so piss easy to solve (I have no idea what exactly AB calc is though so you might not have covered this).

Nice trick malongo, I would've just brute-forced the thing and ended up with a page of mess =p

Malongo

Chile3472 Posts

March 13 2009 06:09 GMT

#22

use sin(A+B)= sinA*cosB+sinB*cosA
since sinPi/4=√2/2 just follow the lines 1 by 1
And the lines where almost all copy paste -_-'

Malongo

Chile3472 Posts

March 13 2009 06:12 GMT

#23

On March 13 2009 15:08 Turbovolver wrote:
LOL, the original problem was written so unclearly, because as far as I can tell the 3 and the -2 have already been pre-multiplied. You should just be adding the three things together, not subbing them in to y'' = 2y' - 3y.

But hell, that ODE is so piss easy to solve (I have no idea what exactly AB calc is though so you might not have covered this).

Nice trick malongo, I would've just brute-forced the thing and ended up with a page of mess =p

Actually he doesnt have to solve the ODE just prove that y satisfy the ODE. I love trigonometry

Hypnosis

United States2061 Posts

March 13 2009 07:24 GMT

#24

isnt y'' the second derivative of y? this is stupid problem that exists to look complicated when its really a stupid problem with no real solving reason... calc isnt hard it just looks hard lol.

Turbovolver

Australia2394 Posts

March 13 2009 07:50 GMT

#25

On March 13 2009 16:24 Hypnosis wrote:
isnt y'' the second derivative of y? this is stupid problem that exists to look complicated when its really a stupid problem with no real solving reason... calc isnt hard it just looks hard lol.

Yep, exactly.

Also Malongo, that trig isn't necessary for the solution, it just makes it a lot cleaner. You probably realise that but whatever.

Bitter

United States2 Posts

March 13 2009 10:13 GMT

#26

well although it should be possible to just work find y' and y'' and then plug it in.... for some reason when i tried that it didn't seem to work... needless to say it is 5 in the morning.

luckily with a little diff e q ...

r^2-2r+3=0
r= 1 +/- √2i

gives the general formula:

y = c1 e^x cos√2x + c2 e^x sin√2x

so as long as the c1 and c2 are correct everything seems right so go ahead and assume for sure that it is in fact cos√2 * x and etc for whoever wants to type everything out. It is simply derivative of first multiplied by second + first times derivative of second over and over and over. now this is annoying to write out so i would make a variable v=sin√2x and w=sin√2x and you end up with

-2v-2w=v''+w''

which is true. lemme know where you are having problems (yay me for first post on teamliquid being a calc post

ninjafetus

United States231 Posts

March 13 2009 19:53 GMT

#27

The ODE works if you just start with y and take the derivatives yourself. The y' and y'' in the OP are incorrect. Just use the product rule and chain rule.

http://www.webpages.ttu.edu/jmillhol/calcproblem.pdf (rotate clockwise to read :p)

Hypnosis

United States2061 Posts

March 13 2009 21:14 GMT

#28

So it's hardly a proof. Unless your trying to prove that something is true using the limit h goes to zero formula..

Cloud

Sexico5880 Posts

March 13 2009 21:24 GMT

#29

Oh well youre not derivating. :/

ninjafetus

United States231 Posts

March 14 2009 07:44 GMT

#30

On March 14 2009 06:14 Hypnosis wrote:
So it's hardly a proof. Unless your trying to prove that something is true using the limit h goes to zero formula..

Well, it depends on what the question is. Since the question was "show that this y satisfied this ODE", you could call it a proof to take the derivatives and plug it in. It doesn't need deltas and epsilons or some fancy theoretical framework to be called a proof; it just needs to verify some statement. Besides, you could argue the "limit as h goes to zero formula" is not a rigorous proof either, since you haven't rigorously justified the limit. What it comes down to is the framework you're expected to solve the problem in.

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