- #1
VinnyCee
- 489
- 0
Here is the problem:
[tex]X'\,=\,\left(\begin{array}{cc}-2 & 5 \\-2 & 4 \end{array}\right)\,X,\,\,\,\,X(0)\,=\,\left(\begin{array}{c} 1 \\ 0 \end{array}\right)[/tex]
Here is what I have:
[tex]det(A\,-\,r\,I)\,=\,0[/tex]
[tex]r^2\,-\,2r\,+\,2\,=\,0[/tex]
[tex]r\,=\,-\frac{1}{2}\,\pm\,i,\,\,\,\,\lambda\,=\,-\frac{1}{2},\,\,\mu\,=\,1[/tex]
This is where I get confused. I have two possibilities for the r-value, right? Namely, they are [itex]r_1\,=\,-\frac{1}{2}\,+\,i[/itex] and [itex]r_2\,=\,-\frac{1}{2}\,-\,i[/itex]. Now I go back and use the [itex]A\,-\,r_n\,I\,=\,0[/itex] equation to figure a general solution for the system.
For [itex]r_1[/itex]:
[tex]\left[A\,-\,(-\frac{1}{2}\,+\,i)\,I\right]\,\left(\begin{array}{c}\xi_1 \\\xi_2 \end{array}\right)\,=\,0[/tex]
[tex]\left(\begin{array}{cc}-\frac{3}{2}\,-\,i & 5 \\-2 & \frac{9}{2}\,-\,i \end{array}\right)\,\left(\begin{array}{c}\xi_1 \\\xi_2 \end{array}\right)\,=\,0[/tex]
[tex]\left(-\frac{3}{2}\,-\,i\right)\,\xi_1\,+\,5\,\xi_2\,=\,0[/tex]
[tex]5\,\xi_2\,=\,\left(\frac{3}{2}\,+\,i\right)\,\xi_1[/tex]
[tex]\xi_2\,=\,\left(\frac{3}{10}\,+\,\frac{1}{5}\,i\right)\,\xi_1[/tex]
[tex]\xi^{(1)}\,=\,\left(\begin{array}{c}\ 1 \\\ \frac{3}{10}\,+\,\frac{1}{5}\,i \end{array}\right)[/tex]
[tex]\xi^{(1)}\,=\,\left(\begin{array}{c}\ 1 \\\ \frac{3}{10} \end{array}\right)\,+\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,i[/tex]
Now, finally, here is the general solution for [itex]r_1[/itex]:
[tex]X_1_c\,=\,C_1\,e^{-\frac{t}{2}}\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{10} \end{array}\right)\,cos\,t\,-\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,sin\,t\right]\,+\,C_2\,e^{-\frac{t}{2}}\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{10} \end{array}\right)\,sin\,t\,+\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,cos\,t\right][/tex]
Using the initial condition given to solve for c1 and c2, I get:
[tex]C_1\,=\,1,\,\,\,\,\,\,C_2\,=\,-\frac{3}{2}[/tex]
Plugging back into the general EQ:
[tex]X_1\,=e^{-\frac{t}{2}}\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{10} \end{array}\right)\,cos\,t\,-\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,sin\,t\right]\,+\,\left(-\frac{3}{2}\right)\,e^{-\frac{t}{2}}\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{10} \end{array}\right)\,sin\,t\,+\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,cos\,t\right][/tex]
And simplifying:
[tex]X_1\,=\,e^{-\frac{t}{2}}\,\left(\begin{array}{c}\ cos\,t\,-\,\frac{3}{2}\,sin\,t \\\ -\frac{13}{20}\,sin\,t \end{array}\right)[/tex]
Now here is where I am confused. I get this answer for [itex]X_1[/itex], and I get a similar one for [itex]X_2[/itex] using the [itex]r_2[/itex] complex conjugate and the initial values:
[tex]X_2\,=\,e^{-\frac{t}{2}}\,\left(\begin{array}{c}\ -2\,sin\,t \\\ -\frac{3}{5}\,sin\,t\,-\,\frac{2}{5}\,cos\,t \end{array}\right)[/tex]
Which one is correct (if either)? Or are they somehow the same?
[tex]X'\,=\,\left(\begin{array}{cc}-2 & 5 \\-2 & 4 \end{array}\right)\,X,\,\,\,\,X(0)\,=\,\left(\begin{array}{c} 1 \\ 0 \end{array}\right)[/tex]
Here is what I have:
[tex]det(A\,-\,r\,I)\,=\,0[/tex]
[tex]r^2\,-\,2r\,+\,2\,=\,0[/tex]
[tex]r\,=\,-\frac{1}{2}\,\pm\,i,\,\,\,\,\lambda\,=\,-\frac{1}{2},\,\,\mu\,=\,1[/tex]
This is where I get confused. I have two possibilities for the r-value, right? Namely, they are [itex]r_1\,=\,-\frac{1}{2}\,+\,i[/itex] and [itex]r_2\,=\,-\frac{1}{2}\,-\,i[/itex]. Now I go back and use the [itex]A\,-\,r_n\,I\,=\,0[/itex] equation to figure a general solution for the system.
For [itex]r_1[/itex]:
[tex]\left[A\,-\,(-\frac{1}{2}\,+\,i)\,I\right]\,\left(\begin{array}{c}\xi_1 \\\xi_2 \end{array}\right)\,=\,0[/tex]
[tex]\left(\begin{array}{cc}-\frac{3}{2}\,-\,i & 5 \\-2 & \frac{9}{2}\,-\,i \end{array}\right)\,\left(\begin{array}{c}\xi_1 \\\xi_2 \end{array}\right)\,=\,0[/tex]
[tex]\left(-\frac{3}{2}\,-\,i\right)\,\xi_1\,+\,5\,\xi_2\,=\,0[/tex]
[tex]5\,\xi_2\,=\,\left(\frac{3}{2}\,+\,i\right)\,\xi_1[/tex]
[tex]\xi_2\,=\,\left(\frac{3}{10}\,+\,\frac{1}{5}\,i\right)\,\xi_1[/tex]
[tex]\xi^{(1)}\,=\,\left(\begin{array}{c}\ 1 \\\ \frac{3}{10}\,+\,\frac{1}{5}\,i \end{array}\right)[/tex]
[tex]\xi^{(1)}\,=\,\left(\begin{array}{c}\ 1 \\\ \frac{3}{10} \end{array}\right)\,+\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,i[/tex]
Now, finally, here is the general solution for [itex]r_1[/itex]:
[tex]X_1_c\,=\,C_1\,e^{-\frac{t}{2}}\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{10} \end{array}\right)\,cos\,t\,-\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,sin\,t\right]\,+\,C_2\,e^{-\frac{t}{2}}\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{10} \end{array}\right)\,sin\,t\,+\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,cos\,t\right][/tex]
Using the initial condition given to solve for c1 and c2, I get:
[tex]C_1\,=\,1,\,\,\,\,\,\,C_2\,=\,-\frac{3}{2}[/tex]
Plugging back into the general EQ:
[tex]X_1\,=e^{-\frac{t}{2}}\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{10} \end{array}\right)\,cos\,t\,-\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,sin\,t\right]\,+\,\left(-\frac{3}{2}\right)\,e^{-\frac{t}{2}}\,\left[\left(\begin{array}{c}\ 1 \\\ \frac{3}{10} \end{array}\right)\,sin\,t\,+\,\left(\begin{array}{c}\ 0 \\\ \frac{1}{5} \end{array}\right)\,cos\,t\right][/tex]
And simplifying:
[tex]X_1\,=\,e^{-\frac{t}{2}}\,\left(\begin{array}{c}\ cos\,t\,-\,\frac{3}{2}\,sin\,t \\\ -\frac{13}{20}\,sin\,t \end{array}\right)[/tex]
Now here is where I am confused. I get this answer for [itex]X_1[/itex], and I get a similar one for [itex]X_2[/itex] using the [itex]r_2[/itex] complex conjugate and the initial values:
[tex]X_2\,=\,e^{-\frac{t}{2}}\,\left(\begin{array}{c}\ -2\,sin\,t \\\ -\frac{3}{5}\,sin\,t\,-\,\frac{2}{5}\,cos\,t \end{array}\right)[/tex]
Which one is correct (if either)? Or are they somehow the same?
Last edited: