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4. (Fall 2016 Midterm) Suppose Romeo and Juliet's love affair is modeled with the following system of differential equations: $\dot{R} = aR + bJ$ $\dot{J} = cR + dJ$ where R(t) denotes how much Romeo loves Juliet, J(t) denotes how much Juliet loves Romeo, and a, b, c, d are constants. Let $x = \begin{pmatrix} R \ J \end{pmatrix}$ be the state vector. Then the system can be represented in matrix form as $\dot{x} = Ax = \begin{pmatrix} a & b \ c & d \end{pmatrix} x$. Let $a = d = -1$ and $b = c = 1$. (a) Determine the eigenvalues and eigenvectors of A. (b) Determine $\Phi(t) = e^{At}$. (c) Let the initial condition be given as $x(0) = \begin{pmatrix} 10 \ 0 \end{pmatrix}$. This means that at $t = 0$, Romeo loves Juliet a lot, but Juliet feels neutral towards Romeo. Find the solution $x(t)$. Is the system stable? What are the values of R(t) and J(t) as $t \to \infty$?

          4. (Fall 2016 Midterm) Suppose Romeo and Juliet's love affair is modeled with the following system
of differential equations:
$\dot{R} = aR + bJ$
$\dot{J} = cR + dJ$
where R(t) denotes how much Romeo loves Juliet, J(t) denotes how much Juliet loves Romeo, and
a, b, c, d are constants. Let $x = \begin{pmatrix} R \ J \end{pmatrix}$ be the state vector. Then the system can be represented in
matrix form as $\dot{x} = Ax = \begin{pmatrix} a & b \ c & d \end{pmatrix} x$. Let $a = d = -1$ and $b = c = 1$.
(a) Determine the eigenvalues and eigenvectors of A.
(b) Determine $\Phi(t) = e^{At}$.
(c) Let the initial condition be given as $x(0) = \begin{pmatrix} 10 \ 0 \end{pmatrix}$. This means that at $t = 0$, Romeo loves
Juliet a lot, but Juliet feels neutral towards Romeo. Find the solution $x(t)$. Is the system stable?
What are the values of R(t) and J(t) as $t \to \infty$?
        
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4. (Fall 2016 Midterm) Suppose Romeo and Juliet's love affair is modeled with the following system
of differential equations:
Ṙ = aR + bJ
J̇ = cR + dJ
where R(t) denotes how much Romeo loves Juliet, J(t) denotes how much Juliet loves Romeo, and
a, b, c, d are constants. Let x = 
    < p m a t r i x > be the state vector. Then the system can be represented in
matrix form as ẋ = Ax = 
    < p m a t r i x >
 x. Let a = d = -1 and b = c = 1.
(a) Determine the eigenvalues and eigenvectors of A.
(b) Determine Φ(t) = e^At.
(c) Let the initial condition be given as x(0) = 
    < p m a t r i x >. This means that at t = 0, Romeo loves
Juliet a lot, but Juliet feels neutral towards Romeo. Find the solution x(t). Is the system stable?
What are the values of R(t) and J(t) as t →∞?

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(Fall 2016 Midterm) Suppose Romeo and Juliet's love affair is modeled with the following system of differential equations: R = aR + bJ J = cR + dJ where Rt denotes how much Romeo loves Juliet, Jt denotes how much Juliet loves Romeo, and a, b, c, and d are constants. Let x be the state vector. Then the system can be represented in matrix form as x' = Ax, where A is a matrix and x' is the derivative of x. Let a = -1 and b = c = 1. a) Determine the eigenvalues and eigenvectors of A. b) Determine A transpose. c) Let the initial condition be given as x0 = [0; 0]. This means that at t = 0, Romeo loves Juliet a lot, but Juliet feels neutral towards Romeo. Find the solution x(t). Is the system stable? What are the values of Rt and Jt as t approaches infinity?
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Transcript

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00:01 So we're given this whole backstory with romeo and juliet.
00:06 And then finally given these differential equations, which govern their love for one another, their love or hatred of one another.
00:17 And said they, first they ask us, why does this agree with the story? well, because we're told that juliet's love grows, the more romeo loves her.
00:26 So this is juliet's love.
00:28 So this is the rate of change of juliet's love.
00:31 This is romeo's love.
00:32 So the more he loves her, the bigger this gets, the bigger this gets.
00:37 And then romeo's love declines, the more juliet loves him.
00:41 Maybe he doesn't like to be...
00:43 He likes his freedom.
00:46 Anyway, we have romeo's love.
00:49 The rate of romeo's love, how it changes is then minus.
00:54 As this grows, this goes down.
00:58 So that's why we have the minus sign there...
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