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In Exercises 9 and 18,construct the general solution of\(x' = Ax\)involving complex Eigen functions and then obtain the general real solution. Describe the shapes of typical trajectories.

13. \(A = \left( {\begin{aligned}{ {20}{c}}4&{ - 3}\\6&{ - 2}\end{aligned}} \right)\)

Short Answer

Expert verified

The requiredcomplex solution is:

\(x(t) = {c_1}\left( {\begin{aligned}{ {20}{c}}{1 + i}\\2\end{aligned}} \right){e^{(1 + 3i)t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{1 - i}\\2\end{aligned}} \right){e^{(1 - 3i)t}}\)

And, the real general solution of is:

\(y(t) = {c_1}\left( {\begin{aligned}{ {20}{c}}{\cos 3t - \sin 3t}\\{2\cos 3t}\end{aligned}} \right){e^t} + {c_2}\left( {\begin{aligned}{ {20}{c}}{\cos 3t + \sin 3t}\\{2\sin 3t}\end{aligned}} \right){e^t}\)

Step by step solution

01

System of Differential Equations

The general solutionfor any system of differential equations withthe eigenvalues\({\lambda _1}\)and\({\lambda _2}\)with the respective eigenvectors\({v_1}\)and\({v_2}\)is given by:

\(x(t) = {c_1}{v_1}{e^{{\lambda _1}t}} + {c_2}{v_2}{e^{{\lambda _2}t}}\)

Here, \({c_1}\) and \({c_2}\) are the constants from the initial condition.

02

Calculation for eigenvalues

It is given that,\(A = \left( {\begin{aligned}{ {20}{l}}4&{ - 3}\\6&{ - 2}\end{aligned}} \right)\).

For eigenvalues, we have:

\(\begin{aligned}{c}\det \left( {x{I_2} - A} \right) = 0\\(4 - x)( - 2 - x) + 18 = 0\\{x^2} - 2x + 12 = 0\\{x_1} = 1 + 3i,{\rm{ }}{x_2} = 1 - 3i\end{aligned}\)

03

Find eigenvectors for both eigenvalues

Now, forthe eigenvalue\(1 + 3i\), we have:

\(\left( {\begin{aligned}{ {20}{c}}{3 - 3i}&{ - 3}\\6&{ - 3 - 3i}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{l}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{l}}0\\0\end{aligned}} \right)\)

Taking\({x_2} = 1\), we get, eigenvector:

\({v_1} = \left( {\begin{aligned}{ {20}{c}}{1 + i}\\2\end{aligned}} \right)\)

Similarly,forthe eigenvalue\(1 - 3i\), we have:

\({v_2} = \left( {\begin{aligned}{ {20}{c}}{1 - i}\\2\end{aligned}} \right)\)

Using both eigenvectors, we have:

\(\begin{aligned}{c}x(t) = {c_1}{v_1}{e^{{\lambda _1}t}} + {c_2}{v_2}{e^{{\lambda _2}t}}\\ = {c_1}\left( {\begin{aligned}{ {20}{c}}{1 + i}\\2\end{aligned}} \right){e^{(1 + 3i)t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{1 - i}\\2\end{aligned}} \right){e^{(1 - 3i)t}}\end{aligned}\)

Hence, this is the required solution.

04

Find the real general solution

Now, the real general solution will be given as:

\(\begin{aligned}{c}x(t) = {c_1}\left( {\begin{aligned}{ {20}{c}}{1 + i}\\2\end{aligned}} \right){e^{(1 + 3i)t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{1 - i}\\2\end{aligned}} \right){e^{(1 - 3i)t}}\\ = {c_1}\left( {\begin{aligned}{ {20}{c}}{1 + i}\\2\end{aligned}} \right)(\cos 3t + i\sin 3t){e^t} + {c_2}\left( {\begin{aligned}{ {20}{c}}{1 - i}\\2\end{aligned}} \right)(\cos 3t - i\sin 3t){e^t}\\ = {c_1}\left( {\begin{aligned}{ {20}{c}}{\cos 3t - \sin 3t}\\{2\cos 3t}\end{aligned}} \right){e^t} + {c_2}\left( {\begin{aligned}{ {20}{c}}{\cos 3t + \sin 3t}\\{2\sin 3t}\end{aligned}} \right){e^t}\end{aligned}\)

Hence, the real general solution of\(x' = Ax\)is given by;

\(y(t) = {c_1}\left( {\begin{aligned}{ {20}{c}}{\cos 3t - \sin 3t}\\{2\cos 3t}\end{aligned}} \right){e^t} + {c_2}\left( {\begin{aligned}{ {20}{c}}{\cos 3t + \sin 3t}\\{2\sin 3t}\end{aligned}} \right){e^t}\), where\({c_1},{c_2} \in \mathbb{R}\)

Hence, this is the required general real solution. And, the trajectories are spiral as the eigenvalues are complex numbers.

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Most popular questions from this chapter

Suppose \(A\) is diagonalizable and \(p\left( t \right)\) is the characteristic polynomial of \(A\). Define \(p\left( A \right)\) as in Exercise 5, and show that \(p\left( A \right)\) is the zero matrix. This fact, which is also true for any square matrix, is called the Cayley-Hamilton theorem.

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

2. \({\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,\,\,{\mathop{\rm and}\nolimits} \,\,\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}\)

Question 18: It can be shown that the algebraic multiplicity of an eigenvalue \(\lambda \) is always greater than or equal to the dimension of the eigenspace corresponding to \(\lambda \). Find \(h\) in the matrix \(A\) below such that the eigenspace for \(\lambda = 5\) is two-dimensional:

\[A = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&3&h&0\\0&0&5&4\\0&0&0&1\end{array}} \right]\]

Use Exercise 12 to find the eigenvalues of the matrices in Exercises 13 and 14.

14. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{5}}&{{\bf{ - 6}}}&{{\bf{ - 7}}}\\{\bf{2}}&{\bf{4}}&{\bf{5}}&{\bf{2}}\\{\bf{0}}&{\bf{0}}&{{\bf{ - 7}}}&{{\bf{ - 4}}}\\{\bf{0}}&{\bf{0}}&{\bf{3}}&{\bf{1}}\end{array}} \right)\)

Question: For the matrices in Exercises 15-17, list the eigenvalues, repeated according to their multiplicities.

17. \(\left[ {\begin{array}{*{20}{c}}3&0&0&0&0\\- 5&1&0&0&0\\3&8&0&0&0\\0&- 7&2&1&0\\- 4&1&9&- 2&3\end{array}} \right]\)

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