Question #102651

Determine the Laplace transform of

f(t)=t^2e^2t

Expert's answer

Answer:


We are given: f(t)=t2e2tf(t) = t^2e^{2t}


We need to find out the Laplace transform of f(t)=t2e2tf(t)=t^2e^{2t}


Apply Laplace transformation rule, where f(t)=e2t,k=2f(t)=e^{2t},k=2


We need to compute d2ds2(1s−1)\frac{d^2}{ds^2}(\frac{1}{s-1})


Solve by using the chain rule

dds(1s−1)=−1(s−2)2dds(s−2)=−1(s−2)2\frac{d}{ds}(\frac{1}{s-1})=-\frac{1}{(s-2)^2}\frac{d}{ds}(s-2)=-\frac{1}{(s-2)^2}


Solve by using the chain rule

d2ds2(1s−2)=d2ds2(−1(s−2)2)=2(s−2)2dds(s−2)=2(s−2)3\frac{d^2}{ds^2}(\frac{1}{s-2})=\frac{d^2}{ds^2}(-\frac{1}{(s-2)^2})=\frac{2}{(s-2)^2}\frac{d}{ds}(s-2)=\frac{2}{(s-2)^3}



Hence L{t2e2t}=2(s−2)3\fbox{L\{$t^2e^{2t}$\}=$\frac{2}{(s-2)^3}$}



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