Question

Method of elimination problem! Please help!

Answer 1

can you help me solve x' = 5x + 7y, y' = 2x + 10y for x(t) and y(t) using the method of elimination? I have the initial conditions x(0) = 1 and y(0) = 5, but I cannot find my coeff. A and B correctly.

Answer 2

okay could you please thell me what the thing next to the x is? is it a 1?

Answer 3

It's a prime---dx and dy sort of thing

Answer 4

hi! Please help!
I can was able to get y(t) = Ae^(39/2)t + Be^(21/2)t, I can't get the correct x, and this y version is probably wrong :C

Answer 5

Hello?

Answer 6

Cheering on Turing.. i don't remember how to do this :/

Answer 7

solve the first one for y\[y=\frac17x'-\frac57x\]take the derivative and notice we have an expression for y' that we can use\[y'=\frac17x''-\frac57x'=2x+10y=2x+\frac{10}7x'-\frac{50}7x\]we can now subtract the far right equation and we have a nice 2nd order homogeneous DE\[x''-15x'+36x=0\]the solution to which is\[x=Ae^{-3t}+Be^{-12t}\]now we should be able to get y without much trouble by using the first equation I posted...

Answer 8

\[y=\frac17(x'-5x)=\frac17(-3Ae^{-3t}-12Be^{-12t}-5Ae^{-3t}-5Be^{-12t})\]\[y=-\frac87Ae^{-3t}-\frac{17}7Be^{-12t}\]now we can use the initial condition so find A and B...

Answer 9

\[y(0)=-\frac87A-\frac{17}7B=5\]\[x(0)=A+B=1\]solve the system to get A and B

Answer 10

I just couldn't divide on a calculator right. It's been one of those days. I see now how my two roots were wrong.

Answer 11

\[-\frac97B=\frac{43}7\to B=\frac{43}9\]\[\frac97A=\frac{52}7\to A=\frac{52}9\]I just did this off the top of my head, so it may have an arithmetic error as well
I hate arithmetic, foils all my grand schemes too :P

Answer 12

Okay, well I have B = (43/7) and A = 1-B = (50/7), very close. I am going to enter this and see if it is right. Thank you very much for your help! Virtual hug :)

Answer 13

assuming I did the rest right, wolfram has my answer for the system, though I forgot to make B negative as my work would indicate
http://www.wolframalpha.com/input/?i=solve%20-8x%2F7-17y%2F7%3D5%2Cx%2By%3D1&t=crmtb01
(x and y are A and B respectively)

Answer 14

Hmmmm....I am re-checking my work, because what I entered wasn't right

Answer 15

Yes my constant is a little off
try it with negative B that I have above

Answer 16

Yeah, I think my work was wrong; I will try yours with the - B

Answer 17

Nope, it's not liking my answer... I have put in x(t) = (52/9)e^(-3t) + (-43/9)e^(-12t); and y(t) = (-8/7)(52/9)e^(-3t) + (-17/7)(-43/9)e^(-12t)

Answer 18

oh crap, how did all the exponent get negative lol?
they should be positive
wow, that's a big typo, but should be easy to fix :D

Answer 19

I put negative exponents through the whole dang problem, I'm trippin'

Answer 20

Thanks for noticing that---I have been doing these problems but when I saw the neg sign, I though I had been doing the wrong thing all along; also: talking the neg sign away don't make the answer correct.

Answer 21

the characteristic polynomial I had was\[r^2-15r+36=0\]so yeah, positive exponents
yeah that's gonna change what we got for y (because we based that on x', which brought negative numbers out) so we have to take it from finding y again with the correct\[x=Ae^{3t}+Be^{12t}\]

Answer 22

sorry

Answer 23

oh, and that will change the signs when you took x', right?

Answer 24

So does -8/7 become 32/7? and so on?

Answer 25

I mean -32/7

Answer 26

\[y=\frac17(x'-5x)=-\frac27Ae^{3t}+Be^{12t}\]please check my work on that as I continue...

Answer 27

I have y = (1/7)(3Ae^(3t) + 12Be^(12t)) -5(Ae^(3t) + Be^(3t))

Answer 28

Where did you get your 2/7 from?

Answer 29

I have now A = -12/55, B = 67/55

Answer 30

yes, that is what I have
just simplify the y and you get what I got
I have a different A and B though

Answer 31

\[A+B=1\]\[-\frac27A+B=5\]

Answer 32

\[A=-\frac{28}9\]\[B=\frac{37}9\]is what I got
jesus I hope that's right...

Answer 33

Oh..okay let me try again. What I did above, that was completely wrong, sorry

Answer 34

Okay so it looks like you did not muliply anything to the B?

Answer 35

typo, I mean\[x=-\frac{28}9e^{3t}+\frac{37}9e^{12t}\]\[y=\frac{56}{63}e^{3t}+\frac{37}9e^{12t}\]

Answer 36

What did you multiply A and B by in the y eqn?

Answer 37

A gets multiplied by -2/7
look at the expression for y a few posts up

Answer 38

Okay...last try...I am going to try it unless anyone has any objections!

Answer 39

@tur looks good. you can tweak the solution by reducing 56/63 to 8/9

Answer 40

haha, I hate arithmetic
If phi thinks it looks good I have hope

Answer 41

okedoke here I go

Answer 42

YESSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS. total win. You are awesome :)

Answer 43

sweetness :D

Answer 44

I am going to go over your work again later, since I made so many mistakes in arithmetic. But thanks so much for helping me!

Answer 45

that was a close one, I have to pay more attention I almost messed you up
happy to help, and you're welcome !