Question

12. mathcal{L}^{-1} left{ frac{s - 1}{s^2 + 2s + 5} ight} (t) is equal to A. e^t left( cos(4t) - frac{1}{2} sin(4t) ight) B. e^{-t} left( cos(4t) - frac{1}{2} sin(4t) ight) C. e^{-t} left( cos(2t) - 2 sin(2t) ight) D. e^t left( cos(2t) - sin(2t) ight) E. e^{-t} left( cos(2t) - sin(2t) ight) 13. mathcal{L}^{-1} left{ frac{s e^{-2s}}{s^2 + pi^2} ight} (t) is equal to A. frac{1}{pi} u_2(t) sin(pi t) B. u_2(t) cos(pi t) C. u_pi(t) cos(2t) D. frac{1}{pi} u_2(t) cos(pi t) E. none of the above 14. Given that the matrix A = egin{bmatrix} 2 & 3 \ -3 & 2 end{bmatrix} has a complex eigenvalue 2 + 3i, and a corresponding eigenvector egin{bmatrix} 1 \ i end{bmatrix}, the initial value problem vec{Y}' = A vec{Y}, quad vec{Y}(0) = egin{bmatrix} 1 \ 2 end{bmatrix} has solution A. egin{bmatrix} e^{2t} (cos(3t) - 2 sin(3t)) \ e^{2t} (2 cos(3t) - sin(3t)) end{bmatrix} B. egin{bmatrix} e^t (cos(3t) - 2 sin(3t)) \ e^t (2 cos(3t) + sin(3t)) end{bmatrix} C. egin{bmatrix} e^t (cos(3t) + sin(3t)) \ e^t (2 cos(3t) - 2 sin(3t)) end{bmatrix} D. egin{bmatrix} e^{2t} (cos(3t) + 2 sin(3t)) \ e^{2t} (2 cos(3t) - sin(3t)) end{bmatrix} E. egin{bmatrix} e^{-t} (cos(3t) - sin(3t)) \ e^{-t} (2 cos(3t) - 2 sin(3t)) end{bmatrix}

          12. mathcal{L}^{-1} left{ frac{s - 1}{s^2 + 2s + 5} 
ight} (t) is equal to
A. e^t left( cos(4t) - frac{1}{2} sin(4t) 
ight) B. e^{-t} left( cos(4t) - frac{1}{2} sin(4t) 
ight)
C. e^{-t} left( cos(2t) - 2 sin(2t) 
ight) D. e^t left( cos(2t) - sin(2t) 
ight)
E. e^{-t} left( cos(2t) - sin(2t) 
ight)

13. mathcal{L}^{-1} left{ frac{s e^{-2s}}{s^2 + pi^2} 
ight} (t) is equal to
A. frac{1}{pi} u_2(t) sin(pi t) B. u_2(t) cos(pi t)
C. u_pi(t) cos(2t) D. frac{1}{pi} u_2(t) cos(pi t)
E. none of the above

14. Given that the matrix A = egin{bmatrix} 2 & 3 \ -3 & 2 end{bmatrix} has a complex eigenvalue 2 + 3i, and a
corresponding eigenvector egin{bmatrix} 1 \ i end{bmatrix}, the initial value problem
vec{Y}' = A vec{Y}, quad vec{Y}(0) = egin{bmatrix} 1 \ 2 end{bmatrix}
has solution
A. egin{bmatrix} e^{2t} (cos(3t) - 2 sin(3t)) \ e^{2t} (2 cos(3t) - sin(3t)) end{bmatrix} B. egin{bmatrix} e^t (cos(3t) - 2 sin(3t)) \ e^t (2 cos(3t) + sin(3t)) end{bmatrix}
C. egin{bmatrix} e^t (cos(3t) + sin(3t)) \ e^t (2 cos(3t) - 2 sin(3t)) end{bmatrix} D. egin{bmatrix} e^{2t} (cos(3t) + 2 sin(3t)) \ e^{2t} (2 cos(3t) - sin(3t)) end{bmatrix}
E. egin{bmatrix} e^{-t} (cos(3t) - sin(3t)) \ e^{-t} (2 cos(3t) - 2 sin(3t)) end{bmatrix}

$12. mathcalL^-1 left fracs - 1s^2 + 2s + 5 ight (t) is equal to A. e^t left( cos(4t) - frac12 sin(4t) ight) B. e^-t left( cos(4t) - frac12 sin(4t) ight) C. e^-t left( cos(2t) - 2 sin(2t) ight) D. e^t left( cos(2t) - sin(2t) ight) E. e^-t left( cos(2t) - sin(2t) ight) 13. mathcalL^-1 left fracs e^-2ss^2 + pi^2 ight (t) is equal to A. frac1pi u2(t) sin(pi t) B. u2(t) cos(pi t) C. upi(t) cos(2t) D. frac1pi u2(t) cos(pi t) E. none of the above 14. Given that the matrix A = eginbmatrix 2 3 -3 2 endbmatrix has a complex eigenvalue 2 + 3i, and a corresponding eigenvector eginbmatrix 1 i endbmatrix, the initial value problem vecY' = A vecY, quad vecY(0) = eginbmatrix 1 2 endbmatrix has solution A. eginbmatrix e^2t (cos(3t) - 2 sin(3t)) e^2t (2 cos(3t) - sin(3t)) endbmatrix B. eginbmatrix e^t (cos(3t) - 2 sin(3t)) e^t (2 cos(3t) + sin(3t)) endbmatrix C. eginbmatrix e^t (cos(3t) + sin(3t)) e^t (2 cos(3t) - 2 sin(3t)) endbmatrix D. eginbmatrix e^2t (cos(3t) + 2 sin(3t)) e^2t (2 cos(3t) - sin(3t)) endbmatrix E. eginbmatrix e^-t (cos(3t) - sin(3t)) e^-t (2 cos(3t) - 2 sin(3t)) endbmatrix$

Added by Leslie H.

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