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The Laplace transform of f(t) = sin πt is $$F\left( s \right) = \frac{\pi }{{{s^2}\left( {{s^2} + {\pi ^2}} \right)}},\,s > 0.$$     Therefore, the Laplace transform of t sin πt is

The Laplace transform of f(t) = sin πt is $$F\left( s \right) = \frac{\pi }{{{s^2}\left( {{s^2} + {\pi ^2}} \right)}},\,s > 0.$$     Therefore, the Laplace transform of t sin πt is Correct Answer $$\frac{{2\pi s}}{{{{\left( {{s^2} + {\pi ^2}} \right)}^2}}}$$

Related Questions

Laplace transform of the function f(t) is given by $${\text{F}}\left( {\text{s}} \right) = {\text{L}}\left\{ {{\text{f}}\left( {\text{t}} \right)} \right\} = \int_0^\infty {{\text{f}}\left( {\text{t}} \right){{\text{e}}^{ - {\text{st}}}}{\text{dt}}{\text{.}}} $$       Laplace transform of the function shown below is given by
Transform Theory mcq question image
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