Find the solution for the following initial value problemSolution.pdf

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For the polynomial below, 2 is a zero. f[x) = x - 8x + 16x - 8 Expres.pdf

For the polynomial below, 2 is a zero. f[x) = x - 8x + 16x - 8 Expres.pdf

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Find the solution for the following initial value problemSolution.pdf

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Find the solution for the following initial value problem Solution The characteristic equation is: r2 + 2r + 5 = 0 r = -1 + 2i , -1 - 2i So the general solution is: y = Ae-tsin(2t) + Be-tcos(2t) y(18) = e-18 (Asin(36) + Bcos(36)) = 6 y\'(18) = e-18 ((-A-2B)sin(36) + (-B+2A)cos(36)) = -7 Therefore: A.sin(36) + B.cos(36) = 6e18 A(2cos(36) - sin(36)) + B(-2sin(36)-cos(36)) = -7e18 Solving this system of equations results in: A = -1/2 [(-2sin(36)-cos(36))*6e18 - cos(36)*(-7e18)] = (6sin(36) - 1/2 cos(36))e18 B = -1/2 [-(2cos(36) - sin(36))6e18 + sin(36)*(-7e18)] = (1/2 sin(36) + 6cos(36))e18 Therefore the solution is: y = (6sin(36) - 1/2 cos(36))e18e-tsin(2t) + (1/2 sin(36) + 6cos(36))e18e-tcos(2t).

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Find the solution for the following initial value problemSolution.pdf

Find the solution for the following initial value problem Solution The characteristic equation is: r2 + 2r + 5 = 0 r = -1 + 2i , -1 - 2i So the general solution is: y = Ae-tsin(2t) + Be-tcos(2t) y(18) = e-18 (Asin(36) + Bcos(36)) = 6 y'(18) = e-18 ((-A-2B)sin(36) + (-B+2A)cos(36)) = -7 Therefore: A.sin(36) + B.cos(36) = 6e18 A(2cos(36) - sin(36)) + B(-2sin(36)-cos(36)) = -7e18 Solving this system of equations results in: A = -1/2 [(-2sin(36)-cos(36))*6e18 - cos(36)*(-7e18)] = (6sin(36) - 1/2 cos(36))e18 B = -1/2 [-(2cos(36) - sin(36))6e18 + sin(36)*(-7e18)] = (1/2 sin(36) + 6cos(36))e18 Therefore the solution is: y = (6sin(36) - 1/2 cos(36))e18e-tsin(2t) + (1/2 sin(36) + 6cos(36))e18e-tcos(2t)