Step 1
Find
\(L^{-1}\left\{\frac{2s+3}{(s-7)^4}\right\}\)
For this take partial fraction
\(\frac{2s+3}{(s-7)^4}=\frac{2}{(s-7)^3}+\frac{17}{(s-7)^4}\)
Step 2
Use linearity property of Inverse Laplace transformation
That is , for function f(s),g(s) and a,b be any constants.
\(L^{-1}\left\{af(s)+bf(s)\right\}=aL^{-1}\left\{f(s)\right\}+bL^{-1}\left\{g(s)\right\}\)
Step 3
Now equation (i) becomes
\(L^{-1}\left\{\frac{2s+3}{(s-7)^4}\right\}=L^{-1}\left\{\frac{2}{(s-7)^3}\right\}+L^{-1}\left\{\frac{17}{(s-7)^4}\right\} =e^{7t}t^2+\frac{17e^{7t}t^3}{6}\)
Formula used
\(L^{-1}(F(s))=f(t)\)
\(L^{-1}(F(S-a))=e^{at}f(t)\)
\(L^{-1}\left(\frac{n!}{S^{n+1}}\right)=t^n\)
Question
Find the inverse of Laplace transform frac{2s+3}{(s-7)^4}
Answers (1)
2020-12-31
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