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The Laplace transform of the function f (t) = et2 – 4t will be—
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1 + e–4s e– t s2 2s -
e–4s2 s3 -
e–4s s3 - does not exist
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Correct Option: D
The Laplace transform is given by the relation for any function f (t).
| L {f(t)} = |  | ∞ | f (t) e–st dt | 0 |
Provided that the integral must be exist within the limit.
But here,
| L {f(t)} = | | ∞ | et2 – 4t. e–st dt | 0 |
