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Inverse laplace transform of ๐ฟโˆ’11๐‘ 2+๐‘Ž2L โˆ’1 s 2 +a 2 1โ€‹ =a)1๐‘Ž๐‘๐‘œ๐‘ ๐‘Ž๐‘กa1โ€‹ cosatb)1๐‘Ž๐‘ ๐‘–๐‘›๐‘Ž๐‘กa1โ€‹ sinatc)1๐‘Ž๐‘ ๐‘–๐‘›๐‘กa1โ€‹ sintd)1๐‘Ž๐‘๐‘œ๐‘ ๐‘กa1โ€‹ cost

Question

Inverse laplace transform of ๐ฟโˆ’11๐‘ 2+๐‘Ž2L โˆ’1 s 2 +a 2 1โ€‹ =a)1๐‘Ž๐‘๐‘œ๐‘ ๐‘Ž๐‘กa1โ€‹ cosatb)1๐‘Ž๐‘ ๐‘–๐‘›๐‘Ž๐‘กa1โ€‹ sinatc)1๐‘Ž๐‘ ๐‘–๐‘›๐‘กa1โ€‹ sintd)1๐‘Ž๐‘๐‘œ๐‘ ๐‘กa1โ€‹ cost

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Solution

The inverse Laplace transform of 1/(s^2 + a^2) is b) 1/a sin(at). This is a standard result in Laplace transform tables. The function 1/(s^2 + a^2) corresponds to the sine function in the time domain, scaled by 1/a.

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