Solve the initial value problem in
The solution is .
Consider the linear differential operator
The characteristic polynomial of is defined as
The characteristic polynomial of the operator as follows.
Then the characteristic polynomial is as follows.
Solve the characteristic polynomial and find the roots as follows.
Therefore, the roots of the characteristic polynomials are 3 and -3.
Since, the roots of the characteristic equation is different real numbers.
The exponential functions form a basis of the kernel of T.
Hence, they form a basis of the solution space of the homogenous differential equation is .
Thus, the general solution of the differential equation .
Consider the general solution of the with the initial value problem is as follows
role="math" localid="1660809514458"Simplify further.
Hence the solution is .
Consider a noninvertible matrix A with two distinct eigenvalues. (Note that one of the eigenvalue must be 0.) Choose two eigenvectors localid="1659699950165" and with eigenvalues localid="1659700076311" and as shown in the accompanying figures. Suppose is negative. Sketch a phase portrait for the system , clearly indicating the shape and long-term behavior of the trajectories.
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