- #1

zaboda42

- 32

- 0

[tex]x_1' = -5x_1+5x_3[/tex]

[tex]x_2' = 5x_1-2x_2[/tex]

[tex]x_3' = 2x_2-5x_3[/tex]

(A) Initially there are 10 pounds of grain in each tank. What will the amounts be as [tex]t \rightarrow \infty[/tex]?

(B) Solve the system and verify your conclusion from (A).

I'm not really sure how I can say anything about (A), so I would appreciate some help there. Here's how I worked out (B), but I'm not really sure how I can finish to "verify my conclusion from (A)":

[tex]{\bf{x}}' = \left[ \begin{array}{cccc} -5&0&5\\5&-2&0\\0&2&-5 \end{array} \right]\left[ \begin{array}{cccc} x_1\\x_2\\x_3 \end{array} \right][/tex]

[tex]{\bf{A}}-\lambda {\bf{I}} = \left[ \begin{array}{cccc} -5-\lambda&0&5\\5&-2-\lambda&0\\0&2&-5-\lambda \end{array} \right][/tex]

Which gives [tex]-\lambda(\lambda^2+12\lambda+45) = 0[/tex] for solutions [tex]\lambda = 0[/tex] and [tex]\lambda = -6\pm3i[/tex]

In the case of [tex]\lambda = 0[/tex]:

[tex]({\bf{A}}-0{\bf{I}}){\bf{v}} \Rightarrow \left[ \begin{array}{cccc} -5&0&5\\5&-2&0\\0&2&-5 \end{array} \right]\left[ \begin{array}{cccc} a\\b\\c \end{array} \right] = \left[ \begin{array}{cccc} 0\\0\\0 \end{array} \right][/tex]

Which gives the system:

[tex]-5a+5c = 0[/tex]

[tex]5a-2b = 0[/tex]

[tex]2b-5c = 0[/tex]

Thus, [tex]{\bf{x}}_0 = \left[ \begin{array}{cccc} 2\\5\\2 \end{array} \right][/tex].

In the case of [tex]\lambda = -6-3i[/tex]:

[tex]({\bf{A}}-(-6-3i){\bf{I}}){\bf{v}} \Rightarrow \left[ \begin{array}{cccc} 1+3i&0&5\\5&4+3i&0\\0&2&1+3i \end{array} \right]\left[ \begin{array}{cccc} a\\b\\c \end{array} \right] = \left[ \begin{array}{cccc} 0\\0\\0 \end{array} \right][/tex]

Which gives [tex]{\bf{x}} = \left[ \begin{array}{cccc} 5\\-4+3i\\-1-3i \end{array} \right][/tex]

So we have:

[tex]{\bf{x}}(t) = \left[ \begin{array}{cccc} 5\\-4+3i\\-1-ei \end{array} \right]e^{(-6-3i)t}[/tex]

[tex]{\bf{x}}(t) = \left[ \begin{array}{cccc} 5\\-4+3i\\-1-ei \end{array} \right]e^{-6t}(\cos3t-i\sin3t)[/tex]

[tex]{\bf{x}}(t) = e^{-6t}\left[ \begin{array}{cccc} 5\cos3t-5i\sin3t\\3\sin3t-4\cos3t+i(4\sin3t+3\cos3t)\\-3\sin3t-\cos3t+i(\sin3t-3\cos3t) \end{array} \right][/tex]

And the real and imaginary parts of [tex]{\bf{x}}(t)[/tex] are real-valued solutions:

[tex]{\bf{x}}_1(t) = e^{-6t}\left[ \begin{array}{cccc} 5\cos3t\\3\sin3t-4\cos3t\\-3\sin3t-\cos3t \end{array} \right][/tex]

[tex]{\bf{x}}_2(t) = e^{-6t}\left[ \begin{array}{cccc} 5\sin3t\\4\sin3t+4\cos3t\\\sin3t-3\cos3t \end{array} \right][/tex]

So the general solution can be written as:

[tex]{\bf{x}}(t) = c_1\left[ \begin{array}{cccc} 2\\5\\2 \end{array} \right] + c_2e^{-6t}\left[ \begin{array}{cccc} 5\cos3t\\3\sin3t-4\cos3t\\-3\sin3t-\cos3t \end{array} \right] + c_3e^{-6t}\left[ \begin{array}{cccc} 5\sin3t\\4\sin3t+3\cos3t\\\sin3t-3\cos3t \end{array} \right][/tex]

With the initial condition, we have a system:

[tex]2c_1+5c_3 = 10[/tex]

[tex]5c_1+3c_2+4c_3 = 10[/tex]

[tex]2c_1 - 3c_2+c_3 = 10[/tex]

So that [tex]c_1 = 2[/tex], [tex]c_2 = -\frac{8}{5}[/tex], and [tex]c_3 = \frac{6}{5}[/tex] so that our solution becomes:

[tex]{\bf{x}}(t) = 2\left[ \begin{array}{cccc} 2\\5\\2 \end{array} \right] -\frac{8}{5}e^{-6t}\left[ \begin{array}{cccc} 5\cos3t\\3\sin3t-4\cos3t\\-3\sin3t-\cos3t \end{array} \right] + \frac{6}{5}e^{-6t}\left[ \begin{array}{cccc} 5\sin3t\\4\sin3t+3\cos3t\\\sin3t-3\cos3t \end{array} \right][/tex]

I hope I did everything correctly, but what does this tell me about the grain as [tex]t \rightarrow \infty[/tex]?