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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!

Mathematics
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Other answers:

  • phi
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'= -2x-y y'= 9x-2y or \[ \frac{dx}{dt}= -2x-y \] \[ \frac{dy}{dt}= 9x-2y \] or using operator notation (where D is the operation derivative wrt t) (D+2)x + y = 0 9 x + (D+2)y=0
  • phi
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 -4D-4-9)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 phase-shifted 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?
im sorry im confused! can you please finish
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
  • phi
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. ?
so in order to solve for y, you derive x?
  • phi
yes, that is one way. the first equation is x'= -2x-y having found x=exp(-2t)*( A cos(3t) + B sin(3t)) we can now find y
  • phi
once we have x and y, use the initial conditions to solve for the constants A and B
so y=x'-2(exp(-2t)*( A cos(3t) + B sin(3t)) )
  • phi
if you know x, you can find x' right?
right
  • phi
so find x' and plug into so y=x'-2(exp(-2t)*( A cos(3t) + B sin(3t)) )
  • phi
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')
ok i found x'=e^(-2t)((-3A-2B)sin(3t)+(3B-2A)cos(3t))
is that right
  • phi
looks good. now sub into y= -x'-2(exp(-2t)*( A cos(3t) + B sin(3t)) )
ok i did that and ended upo getting A=0.9 and B=31/30
is that what you got as well?
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 :(
  • phi
no, I will have to work through it to get the details.
im not sure what I did wrong because I did as you said and my end result was incorrect im not sure why though
but thanks
  • phi
I am sure it is a simple algebra mistake. start with y= -e^(-2t)((-3A-2B)sin(3t)+(3B-2A)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 \[ y= e^{-2t}(3Asin(3t)-3Bcos(3t)) \]
  • phi
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))
thank you so much! i see where my calculation error was!
  • phi
yes, these problems require a lot of concentration on detail
they sure do!

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