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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!
 one year ago
 one year 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!
 one year ago
 one year ago

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ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
@mahmit2012
 one year 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
 one year 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?
 one year ago

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

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

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

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

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

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
is that right
 one year 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)) )
 one year 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
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
is that what you got as well?
 one year 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 :(
 one year ago

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

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
but thanks
 one year 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 \[ y= e^{2t}(3Asin(3t)3Bcos(3t)) \]
 one year 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))
 one year ago

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

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

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