2. Solve the linear system dx dt dt 3x -2y, 5x y (t), y (0) 30 Solution Applying

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Explanation / Answer

Applying laplace transform to dx/dt = 3x-2y ............sX[s]- x(0) = 3X[s]-2Y[s] ==> sX[s] = 3X[s]-2Y[s] ==>(s-3) X[s]+2Y[s] = 0------(1) ..................Applying laplace transform to dy/dt = 5x+y+?(t) ........sY[s] - y(0) = 5X[s]+Y[s]+1 .(since L[?(t)] = 1)...==> -5X[s] +(s-1)Y[s]= 1-------->(2) .......now multiply eqn (1) with 5 and (2) with (s-3) and then add them i.e. 5*(1) + (s-3)*(2) and solve for Y[s] gives ........[(s-1)(s-3)+10]Y[s] = s-3 ...........Y[s] = (s-3) / (s^2 -4s +13) .......... so Y[s] = (s-3) / [(s-2)^2 +3^2] .............Y[s] = (s-2)/[(s-2)^2 +3^2] - (1/3)*3/ [(s-2)^2 +3^2]................ applying Inverse L.T. we have ......y(t) = (e^2t)cos 3t - (1/3)(e^2t)sin 3t................ from eqn (2) we have 5X[s] = (s-1)Y[s] -1 ...........X[s] = -2 / (s^2-4s+13) = (-2/3)*3 / [(s-2)^2+3^2] ..............so x(t) = (-2/3)(e^2t)sin 3t ......................

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