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Yogurt
United States4258 Posts
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 ![[image loading]](http://i241.photobucket.com/albums/ff119/philopo/equation-1.jpg)   | ||
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Fr33t
United States1128 Posts
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Yogurt
United States4258 Posts
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 | ||
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Fr33t
United States1128 Posts
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Yogurt
United States4258 Posts
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Fr33t
United States1128 Posts
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Cloud
Sexico5880 Posts
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micronesia
United States24745 Posts
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Fr33t
United States1128 Posts
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Yogurt
United States4258 Posts
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 | ||
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Fr33t
United States1128 Posts
(%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... | ||
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Yogurt
United States4258 Posts
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 | ||
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Fr33t
United States1128 Posts
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Yogurt
United States4258 Posts
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 =/ | ||
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Fr33t
United States1128 Posts
(%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)) | ||
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Yogurt
United States4258 Posts
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Fr33t
United States1128 Posts
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Malongo
Chile3472 Posts
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 | ||
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Malongo
Chile3472 Posts
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Fr33t
United States1128 Posts
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. | ||
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Turbovolver
Australia2394 Posts
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 | ||
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Malongo
Chile3472 Posts
since sinPi/4=√2/2 just follow the lines 1 by 1 And the lines where almost all copy paste -_-' | ||
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Malongo
Chile3472 Posts
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  | ||
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Hypnosis
United States2061 Posts
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Turbovolver
Australia2394 Posts
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. | ||
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Bitter
United States2 Posts
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 ). | ||
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ninjafetus
United States231 Posts
http://www.webpages.ttu.edu/jmillhol/calcproblem.pdf (rotate clockwise to read :p) | ||
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Hypnosis
United States2061 Posts
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Cloud
Sexico5880 Posts
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ninjafetus
United States231 Posts
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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