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Question Number 129132 by BHOOPENDRA last updated on 13/Jan/21

Answered by Dwaipayan Shikari last updated on 13/Jan/21

$${\mathcal{L}}^{-1}\left(\frac{6as}{(9s^2+a^2)}\right)=\frac{1}{9}{\mathcal{L}}^{-1}\left(\frac{2\left(\frac{a}{3}\right)s}{(s^2+{\left(\frac{a}{3}\right)}^2)}\right)=\frac{t}{9}sin\left(\frac{at}{3}\right)$$

Commented by BHOOPENDRA last updated on 13/Jan/21

$$thankssir$$

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