#$&*
course Mth 279
4/30 3
Query 25 Differential Equations*********************************************
Question: Determine whether the set of solutions {y_1, y_2, y_3} is linearly independent, where
y_1 = [ e^t, 1]
y_2 = [ e^(-t), 1]
y_3 = [ sinh(t), 0]
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Your solution:
Yes set of solutions is linearly independent. When plugging into matrix and attempting to reduce to echelon form(easy to see right off because we have 3 variables and 3 unknowns and only 2 equations, so more than likely we will have to have so sort of linear combination) we see we have solution y_1+y_2 = 0 and y_3=0.
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Given Solution:
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@& The question is whether it's possible for
c_1 y_1 + c_2 y_2 + c_3 y_3
to equal zero, if not all the constants c_1, c_2 and c_3 are zero.
The matrix [y_1,y_2,y_3] reduces to the form
[1, 0; 0, 1; 0, 0]
indicating that the third row is a linear combination of the first two.
Specifically, half the first row minus half the second is equal to the third.
1/2 y_1 + 1/2 y_2 - y_3 = 0.
So the set is not linearly independent.*@
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Question: Determine whether the set of solutions {y_1, y_2, y_3} is linearly independent, where
y_1 = [ 1, sin^2(t), 0]
y_2 = [ 0, 2 - 2 cos^2(t), -2]
y_3 = [ 1, 0, 1]
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Your solution:
I think I’m confused this problem again has form y_1 + y_3 and y_2 - (1/2)y_3 =0, I thought that because this was of form that linear combinations could be used to get 0 vector this was linearly independent but now I starting to recall that system of equations was only linearly independent if system has unique solution where y_1 = y_2 = y_3 =0.
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Given Solution:
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@& If these vectors form a 3 x 3 matrix which row reduces to the identity, the three are linearly independent.
Equivalently if the exist c_1, c_2, c_3 not all equal to zero such that c_1 y_1 + c_2 y_2 + c_3 y_3 = 0, the system is linearly dependent.
It turns out that y_1 - 1/2 y_2 = y_3, so the latter condition holds.
The matrix reduces to
[1, 0, 0; 0, 0, -1; 0, 0, 0].,
equivalently showing the system to be linearly dependent.*@
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Question: Determine whether there is a matrix P(t) such that
y_1 = [ t^2, 0 ]
y_2 = [ 2t, 1 ]
is a fundamental set of solutions to the equation
y ' = P(t) y.
If so, find such a matrix P(t).
Hint: The matrix psi(t) = [y_1, y_2 ] = [ t^2, 2 t; 0, 1 ] would need to satisfy psi ' (t) = P(t) psi(t).
In standard notation we could write this as follows:
satisfies
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Your solution:
Yes this is fundamental set because determinant is nonzero.
Ok, rally getting confused. I see we are shown that psi’(t) = P(t)psi(t) is a form where we can get to if we know psi(t) and it is a fundamental matrix for the homogeneous linear system y’ = Ay. Then we use psi(t) to create fundamental matrix psi_hat(t) that satisfies condition psi_hat(t)(t_0) = Identity matrix. So we can assume that psi_hat(t) is the solution of the matrix init. value problem (psi_hat)’= A*psi, psi(t_0) = Identity matrix
psi(t) = [ t^2, 2 t; 0, 1]
psi(t)^-1 = [t^-2, -2/t; 0, 1], now I get confused were we are to find Identity matrix at t_0, if this was to work it will work for anything other than 0. So do we give general form which is [t^-2, -2/t; 0, 1], -infinity<=t<0 and 0
So that
P(t) = [t^-2, -2/t; 0, 1], -infinity<=t<0 and 0
????How and why is it that psi(t)’ = 1(is this just understood to be the way it is). Having a hard time comprehending and or keeping everything straight that is being talked about in this Ch?????
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Given Solution:
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@& If
psi ' = P(t) psi
then
psi ' * psi^-1 = P(t) psi psi^-1 = P(t)
so
P(t) = psi ' psi^-1 = [ 2 t, 2; 0, 0 ] * [t^-2, -2 / t; 0, 1 ] = [ 2 t^-1 , -2; 0, 0 ]
Does this matrix satisfy
psi ' = P(t) psi?
I think it works for the first function y_1 but not for y_2. Likely a small computational glitch. Can you identify the glitch?*@
@& I believe that W(t) = -2*@
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Question: If the matrix psi(t) = [y_1, y_2] = [e^t, e^(-t); e^t, - e^(-t)]:
What are the vector functions y_1 and y_2?
Write out the system of two differential equations represented by the equation y ' = P(t) y with P(t) = [0, 1; 1, 0].
Show that y_1 and y_2 are both solutions of the equation y ' = P(t) y with P(t) = [0, 1; 1, 0].
Show that { y_1 , y_2} is a fundamental set for this equation.
Show that the matrix psi(t) is a solution of the matrix equation psi ' = P(t) psi.
Show that the matrix psi(t) is a fundamental matrix for the linear system of equations.
Let psi_hat(t) = [ 2 e^t - e^(-t), e^t + 3 e^(-t); 2 e^t + e^(-t), e^t - 3e^(-t) ].
Find a constant matrix C such that psi_hat(t) = psi(t) * C.
Based on your matrix C, is psi_hat(t) a solution matrix for the system?
Based on your matrix C, is psi_hat(t) a fundamental matrix for the system?
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Your solution:
y_1 = [e^t;e^t]
y_2=[ e^(-t); - e^(-t)]
y ' = P(t) y = [0, 1; 1, 0]*[ y_1, y_2]= [0*e^(-t),e^(-t); e^(-t),0*-e^(-t)]
@& The system would be
y_2 ' = y_1
y_1 ' = y_2
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@& You can substitute y_1 and y_2 into these equations and show that both are satisfied.*@
@& *@
W(t) =
@& I believe that W(t) = -2*@
????I see this information in the book but don’t totally understand what’s going on?????
To show that the matrix psi(t) is a solution of the matrix equation psi ' = P(t) psi, we assume that solutions of y’ = P(t)y are used to form a solution matrix psi(t). So that…..
psi(t)’ = [y_1(t)’, y_2(t)’,….,y_n(t)’]
=[P(t)y_1(t), P(t)y_2(t),…….., P(t)y_n(t)]
= P(t)[y_1(t), y_2(t),…….., y_n(t)]
=P(t)*psi(t)
So psi(t) is itself a solution of the matrix differential equation psi(t)’ = P(t)*psi(t)
???I think my problem is understanding the basics because looking at that it makes sense sort of, I just think it’s the foundation of first order linear systems that I’m having trouble understanding?????????
????Show that the matrix psi(t) is a fundamental matrix for the linear system of equations, so we would simply calculate the wronskian to make sure the determinant of psi(t_0) = non-zero. I’m getting over whelming confusion going on in my brain???
Find a constant matrix C such that psi_hat(t) = psi(t) * C.
[ 2 e^t - e^(-t), e^t + 3 e^(-t); 2 e^t + e^(-t), e^t - 3e^(-t) ] = [e^t, e^(-t); e^t, - e^(-t)]*[2,3;-1,1]
c = [2,3;-1,1]
As to is psi_hat(t) a solution matrix for the system or a fundamental matrix for the system?
psi_hat(t) = [1,4;3,-2]
psi(0) = [1,1;1,-1]
c = [2,1;-1,3]
Where
psi_hat(0) = psi(0)*c, what all this is saying exactly I’m am still unclear. I know how to apply the linear algebra stuff and can follow along out of the book with what’s going on but really don’t understand what’s going on???????
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Given Solution:
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@& psi ' (t) = [e^t, -e^-t); e^t, e^-t ]
The linear system has solutions
y_1 = [e^t; e^t]
and
y_2 = [e^-t, -e^-t].
psi = [y_1, y_2].
If you multiply psi by any invertible constant matrix, all you are doing is forming two linear combinations of y_1 and y_2, one in the first column of your product matrix, one in the second.
Any linear combination of the vectors in the fundamental set is a solution.
The two linear combinations are linearly independent as long as the constant matrix is invertible.*@
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Question: Given the system
y ' = [ 1, 1; 0, -2 ] y
verify that
psi(t) = [ e^t, e^(-2 t); 0, e^(-2 t) ]
is a fundamental matrix for the system.
Find a matrix C such that
psi_hat(t) = psi(t) * C
is a solution matrix satisfying initial condition psi_hat(0) = I, where I is the identity matrix.
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Your solution:
det[psi(t_0)] = W(t_0) = e^-t, verifying that psi(t) is a fundamental matrix.
psi(0) = [ 1, 1; 0, 1]
psi^-1(0) = [ 1, 1; 0, 1]^-1= [1,-1;0,1]
So for psi_hat(0) = I and psi_hat(t) = psi(t) * C we have,
psi_hat(t) = [ e^t, e^(-2 t); 0, e^(-2 t) ]* [1,-1;0,1]
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Given Solution:
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#*&!
@& You're getting it. Takes a little bit to put it together, but you're close.
Check my notes. You might well have additional questions.*@