Solve the linear system d2x/dt2 = 2x + 6y, x(0) = 3, x(0) = 0 d2y/dt2 = x + y, y

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Explanation / Answer

Applying laplace transform to d^2x/dt^2 = 2x+6y .......s^2 X[s] - sx(0) - x'(0) =2X[s]+6Y[s] ....==> (s^2 - 2)X[s] - 6Y[s] =3s ------->(1) ...........Now Applying laplace transform to d^2y/dt^2 = x+y ..........s^2 Y[s] - sy(0) - y'(0) =X[s] + Y[s] ....==> - X[s] +(s^2 -1)Y[s] = s ----->(2)...... from (1) and (2) solving for Y[s] , multiply equation (2) by (s^2 - 2) and then add (1) and (2) ......we get ....{(s^2 -1)(s^2 - 2) - 6 } Y[s] = s(s^2 - 2) +3s .........==> Y[s] = (s^3 +s)/(s^4 - 3s^2 -4) ............Y[s] = s(s^2 + 1) / [(s^2 -4)(s^2 +1)] = s / (s^2 -4) ...................so y(t) = L^-1 [s / (s^2 - 2^2)] ..............y(t) = cosh 2t ..................... Now from (2) we have X[s] = (s^2 - 1)Y[s] -s = (s^2 - 1)[ s / (s^2 -4)] - s = s(s^2 -1 -s^2 +4)/(s^2-4) = 3s/(s^2 -4) ..............so x(t) = 3 L^-1[s/(s^2 -4)] = 3cosh 2t ....................so x(t) = 3cosh 2t and y(t) = cosh 2t

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