Problem 4.Consider the initial value problema. Find the eigenvalue X, an eigenvector ₁, and a generalized eigenvector ₂ for the coefficient matrix of this linear system.λ =c. Solve the original initial value problem.y₁ (t) =Y₂ (t) =ÿ=y(t) = C₁v₁ =_1]b. Find the most general real-valued solution to the linear system of differential equations. Use t as the independent variable in your answers.ÿ, ÿ(0) =+ c₂[1].v₂ =[

The given matrix is y→'=-410-4, y→0=1-1 We have to find the eigenvalues and eigenvectors of the…

Answered: Consider the initial value problem a.…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Suppose A is a 2 × 2 real matrix with an eigenvalue A = 1+ 4i and corresponding eigenvector -1+i Determine a fundamental set (i.e., linearly independent set) of solutions for y' = Aj, where the fundamental set consists entirely of real solutions. Enter your solutions below. Use t as the independent variable in your answers. 1 ÿ1(t) ý2(t) =
Consider the linear system X' = [223] -2 X. 1. Show that the eigenvalues of the given system are complex. 2. Find the general solution of the system.
Let A = c. The image of A is a subspace of choose -3 -1 12 14] -3 0 0 a. A basis for the image of A is }. To submit your answer, you may enter a coordinate vector, such as , or a comma separated list of coordinate vectors, such as ,. d. The geometry or shape of the image of A is choose 0 0 b. The dimension of the image of A is than one correct answer): A. Two of the three columns in rref(A) do not have a leading one (or pivot). OB. Two of the three columns in rref(A) have leading ones (or pivots). OC. Two of the three columns in rref(A) are free variable columns. D. rref(A) has a leading one (or pivot) in every row. OE. The basis we found for the image of A has two vectors. F. rref(A) is the identity matrix. because (select all correct answers -- there may be more (enter "R^n" with “n” being an actual number) because
Consider the second-order equation d 2y/dt 2 + p dy/dt +qy = 0 a. If p=q=1, compute the eigenvalues and eigenvectors of the coefficient matrix. What would be the long-term behavior of the solutions in this case? b. If p=0 and q=2 determine the general solution.
The options for matching the values of c1 and c2 are -4 through 1.
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Consider the linear system dx₁ dt dx₂ dt = 4x1 - 3x₂ = 3x1 + 4x2. 1. Show that upon conversion to the form X'= AX, the eigenvalues of A are complex. 2. Find the general solution of the system.
Suppose A is a 2 × 2 real matrix with an eigenvalue λ = 3 + 5i and corresponding eigenvector i * = (-²+1). v= Determine a fundamental set (i.e., linearly independent set) of solutions for ÿ' = Aỹ, where the fundamental set consists entirely of real solutions. Enter your solutions below. Use t as the independent variable in your answers. (t) = 2(t) = → - →
) Consider the linear system X₁ = Y a. Find the eigenvalues and eigenvectors for the coefficient matrix. 18 v1 = b. Find the real-valued solution to the initial value problem {{{ = -3 -2 5 -3] Use t as the independent variable in your answers. y₁ (t) Y₂ (t) - 3y1 - 2y2, 5y1 + 3y2, y." , and X₂ = V2 y₁ (0) = 9, Y₂ (0) = -10. ||
Suppose that a 2 x 2 matrix A with real entries has an eigenvalue X = 1 + 3i with corresponding 4 i [12] eigenvector 1 + 2i Which of the following is NOT a solution of the linear system y' = Ay? y(t) = et y(t) = et y(t) = et y(t) = et 4 cos(3t) sin(3t) [cos(3t) + 2 sin(3t). 4 cos(3t) + sin(3t) cos(3t) 2 sin (3t) - cos(3t) + 4 sin(3t) 2 cos(3t) + sin(3t) cos (3t) - 4 sin (3t) -2 cos(3t) - sin(3t)
I already know the eigenvalues are -2+6i, and 2-6i, and the eigenvector is [6i,1] , and [-6i,1], im just struggling finding the overall solution
Consider the linear system -3 = 5 a. Find the eigenvalues and eigenvectors for the coefficient matrix. λ₁ = v₁ = = b. Find the real-valued solution to the initial value problem Use t as the independent variable in your answers. y₁ (t) = y2(t) = and ₂ = = = -3y1 - 2y2, y₁ (0) = 0, = 5y1 + 3y2, y2 (0) = 5.

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