The discussion focuses on using the Laplace transform to solve the differential equation x'' + 3x' + 5x = sin(8t) with initial conditions x(0) = 2 and x'(0) = -3. The user has derived the equation (s² + 3s + 5)X(s) = 2s + 3 + [8/(s² + 64)] but is seeking assistance to proceed further. A suggested method to advance is to complete the square on the polynomial (s² + 3s + 5) to simplify the problem.
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sim907 said:Use Laplace to solve x''+3x'+5x=sin(8t)
Initial conditions x(0)= 2 and x'= -3
I have worked it down to (s2+3s+5)X(s)= 2s+3+ [8/(s2+64)] but stuck. Please help