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ranyai12
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Can Someone Please Help!! I tried solving this multiple times but IDK for sure what to do! I attached the problem!
 2 years ago
 2 years ago
ranyai12 Group Title
Can Someone Please Help!! I tried solving this multiple times but IDK for sure what to do! I attached the problem!
 2 years ago
 2 years ago

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ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
@mahmit2012
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I think by "find the solution to the linearization around zero" they mean drop all high order terms from the equations. So the problem becomes x'= 2xy y'= 9x2y or \[ \frac{dx}{dt}= 2xy \] \[ \frac{dy}{dt}= 9x2y \] or using operator notation (where D is the operation derivative wrt t) (D+2)x + y = 0 9 x + (D+2)y=0
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
you can solve by multiplying the top equation by (D+2) and adding the equations (D+2)^2 x  (D+2) y =0 9 x +(D+2) y =0 (D^2 4D49)x=0 the characteristic equation for this homogenous differential equation is D^2 4D 13 = 0 the 2 root are 2± 3i the solution can take a number of forms (exponential, cos/sin, or phaseshifted sin) Picking the sin/cos form x= exp(2t)*( A cos(3t) + B sin(3t)) where A and B are constants determined by the initial conditions. to find y, we can sub in for x in one of the equations, say the first one: y= Dx 2x can you finish?
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
im sorry im confused! can you please finish
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
i need to see the woprk to understand because I originally did the matrix and got those roots then after that i got really confused
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
can you find y given x= exp(2t)*( A cos(3t) + B sin(3t)) and y= Dx 2x where D means take the derivative with respect to t. ?
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
so in order to solve for y, you derive x?
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
yes, that is one way. the first equation is x'= 2xy having found x=exp(2t)*( A cos(3t) + B sin(3t)) we can now find y
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
once we have x and y, use the initial conditions to solve for the constants A and B
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
so y=x'2(exp(2t)*( A cos(3t) + B sin(3t)) )
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
if you know x, you can find x' right?
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
so find x' and plug into so y=x'2(exp(2t)*( A cos(3t) + B sin(3t)) )
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
although I think the equation should be y= x'2(exp(2t)*( A cos(3t) + B sin(3t)) ) (a minus sign in front of x')
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
ok i found x'=e^(2t)((3A2B)sin(3t)+(3B2A)cos(3t))
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
is that right
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
looks good. now sub into y= x'2(exp(2t)*( A cos(3t) + B sin(3t)) )
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
ok i did that and ended upo getting A=0.9 and B=31/30
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
is that what you got as well?
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
I ended up with x=(e^(2t))((0.9cos(3t))+((31/30)sin(3t))) and y=((e^(2t)(143/30)sin(3t))+1.3cos(3t))2(e^(2t)(.9cos(3t)+(31/30)sin(3t))) and it ended up being wrong :(
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
no, I will have to work through it to get the details.
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
im not sure what I did wrong because I did as you said and my end result was incorrect im not sure why though
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
but thanks
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I am sure it is a simple algebra mistake. start with y= e^(2t)((3A2B)sin(3t)+(3B2A)cos(3t))2(exp(2t)*( A cos(3t) + B sin(3t)) ) factor out e^(2t) \[ e^{2t}(3Asin(3t)+2Bsin(3t)3Bcos(3t)+2Acos(3t)2Acos(3t)2Bsin(3t)) \] 2Bsin(3t)2Bsin(3t) and 2Acos(3t)2Acos(3t) cancel, so we get \[ y= e^{2t}(3Asin(3t)3Bcos(3t)) \]
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
using y(0)= 0.5 and x(0)= 0.9 in x= exp(2t)(Acos(3t)+B sin(3t)) > A= 0.9 y= exp(2t)(3Asin(3t)3Bcos(3t))> 3B= 1/2, B= 1/6 plugging in A and B we get x= exp(2t)(0.9cos(3t)+(1/6)*sin(3t)) y= exp(2t)(2.7 sin(3t)  0.5 cos(3t))
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
thank you so much! i see where my calculation error was!
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
yes, these problems require a lot of concentration on detail
 2 years ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
they sure do!
 2 years ago
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