Find the inverse Laplace transform of 2s+1(s+2)2(s−1)2
Question
Find the inverse Laplace transform of 2s+1(s+2)2(s−1)2
Solution
The inverse Laplace transform of a function is a complex process that involves finding the original function from its Laplace transform. Here's how you can find the inverse Laplace transform of the given function step by step:
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First, rewrite the function in a form that can be easily recognized in the Laplace transform table. The given function is 2s/(s+2)^2 + 1/(s-1)^2.
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Now, look up each term in the Laplace transform table. The Laplace transform of t*e^(at) is 1/(s-a)^2 and the Laplace transform of e^(at) is 1/(s-a).
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Therefore, the inverse Laplace transform of 1/(s-a)^2 is t*e^(at) and the inverse Laplace transform of 1/(s-a) is e^(at).
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Apply these transforms to each term in the function. The inverse Laplace transform of 2s/(s+2)^2 is 2te^(-2t) and the inverse Laplace transform of 1/(s-1)^2 is te^t.
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Add these two terms together to get the inverse Laplace transform of the original function: 2te^(-2t) + te^t.
Please note that this is a simplified explanation and the actual process may involve more complex algebra and calculus.
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