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The solution to the initial value problem

x''+2x'+82x=4cos(4t),
x(0)=0 x'(0)=0
is the sum of the steady periodic solution xsp and the transient solution xtr. Find both xsp and xtr.

1 Answer

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Final answer:

The solution to the given initial value problem consists of finding a steady periodic solution and a transient solution for the second-order non-homogeneous differential equation. The steady solution takes the form of a cosine function with certain amplitude and phase shift, while the transient solution involves exponential functions that decay over time.

Step-by-step explanation:

The initial value problem provided is a second-order non-homogeneous differential equation with initial conditions. The solution to this problem comprises two parts: the steady periodic solution, xsp(t), and the transient solution, xtr(t). To find the xsp(t), we look for a solution in the form x(t) = A cos(4t + φ), where A is the amplitude and φ is the phase shift. Since the driving force is 4 cos(4t), A would be determined by the particular form of the equation.

For the transient solution, we solve the homogeneous equation x'' + 2x' + 82x = 0 which typically involves exponential functions. Given the initial conditions x(0) = 0 and x'(0) = 0, we can determine the constants in the transient solution. Over time, the transient solution will decay to zero, leaving only the steady periodic solution. Without explicit formulas for A and φ, we cannot provide the exact amplitude and phase shift required to solve this problem, but the procedure has been described.

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