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

- anonymous

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- anonymous

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- anonymous

@dumbcow

- anonymous

@mahmit2012

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## More 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?

- anonymous

im sorry im confused! can you please finish

- anonymous

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.
?

- anonymous

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

- anonymous

so y=x'-2(exp(-2t)*( A cos(3t) + B sin(3t)) )

- phi

if you know x, you can find x' right?

- anonymous

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')

- anonymous

ok i found x'=e^(-2t)((-3A-2B)sin(3t)+(3B-2A)cos(3t))

- anonymous

is that right

- phi

looks good. now sub into
y= -x'-2(exp(-2t)*( A cos(3t) + B sin(3t)) )

- anonymous

ok i did that and ended upo getting A=0.9 and B=31/30

- anonymous

is that what you got as well?

- anonymous

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.

- anonymous

im not sure what I did wrong because I did as you said and my end result was incorrect im not sure why though

- anonymous

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))

- anonymous

thank you so much! i see where my calculation error was!

- phi

yes, these problems require a lot of concentration on detail

- anonymous

they sure do!

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