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

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phi
 2 years ago
Best ResponseYou've already chosen the best response.1I 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

phi
 2 years ago
Best ResponseYou've already chosen the best response.1you 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?

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

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0i 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
 2 years ago
Best ResponseYou've already chosen the best response.1can 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. ?

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

phi
 2 years ago
Best ResponseYou've already chosen the best response.1yes, 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

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

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

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

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

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

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

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

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

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

ranyai12
 2 years ago
Best ResponseYou've already chosen the best response.0I 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
 2 years ago
Best ResponseYou've already chosen the best response.1no, I will have to work through it to get the details.

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

phi
 2 years ago
Best ResponseYou've already chosen the best response.1I 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 \[ y= e^{2t}(3Asin(3t)3Bcos(3t)) \]

phi
 2 years ago
Best ResponseYou've already chosen the best response.1using 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))

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

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