Finding the inverse Laplace transform after solving a differential equationI wanted to solve y″+9y=sin⁡(3x), where y(0)=0 and y′(0)=0 by doing a Laplace Transform.Taking L[y″+9y]=L[sin⁡(3x)] gives s2F(s)−sy(0)−y′(0)+9F(s)=3s2+9The initial values are both zero, and F(s) can be factored out of the left-hand side, leaving :F(s)=3(s2+9)2 and I couldn't figure out how to inverse transform this, but I did see that it was similar to L−1[2a3(s2+a2)2]=sin⁡(at)−atcos⁡(at) which might be helpfulSo in the end, how do I do L−1[3(s2+9)2]?

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Finding the inverse Laplace transform after solving a differential equationI wanted to solve y″+9y=sin⁡(3x), where y(0)=0 and y′(0)=0 by doing a Laplace Transform.Taking L[y″+9y]=L[sin⁡(3x)] gives s2F(s)−sy(0)−y′(0)+9F(s)=3s2+9The initial values are both zero, and F(s) can be factored out of the left-hand side, leaving :F(s)=3(s2+9)2 and I couldn&#039;t figure out how to inverse transform this, but I did see that it was similar to L−1[2a3(s2+a2)2]=sin⁡(at)−atcos⁡(at)  which might be helpfulSo in the end, how do I do L−1[3(s2+9)2]?

gsmckibbenx7ga · Accepted Answer

Explanation:Well, using the table of selected Laplace transforms we can see that:yα(x)&amp;nbsp;=Ls−1[2α3(s2+α2)2](x)=sin⁡(αx)−αxcos⁡(αx)&amp;nbsp;&amp;nbsp;&amp;nbsp;(1)Now, in your case we have:3(s2+9)2=118⋅2⋅33(s2+32)2&amp;nbsp;&amp;nbsp;&amp;nbsp;(2)

Finding the inverse Laplace transform after solving a

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