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Use Laplace transforms to solve the following initial value problem. x′′+25x=sin6t,x(0)=0,x′(0)=0 x(t)=

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Final Answer:


\(x(t) = (1)/(170)\sin(6t) - (6)/(170) \sin(5t)\)This solution represents the combination of sinusoidal functions that satisfy the given initial value problem.

Explanation:

To solve the given initial value problem using Laplace transforms, first, we apply the Laplace transform to the differential equation
\(x'' + 25x = \sin(6t)\)and the initial conditions
\(x(0) = 0\) and \(x'(0) = 0\).

After rearranging and applying the initial conditions, we derive the transformed equation in terms of the Laplace variable
\(s\),which gives us the equation in terms of
\(X(s)\). Solving for \(X(s)\)yields a partial fraction decomposition.

Through inverse Laplace transform and using trigonometric identities, the solution
\(x(t)\) can be expressed as a combination of sine functions with different frequencies, amplitudes, and phase shifts, corresponding to the coefficients obtained from the partial fraction decomposition.

The final solution for
\(x(t)\) is \(x(t) = (1)/(170)\sin(6t) - (6)/(170) \sin(5t)\). This solution represents the combination of sinusoidal functions that satisfy the given initial value problem.

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