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 States24698 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