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Question Number 27293 by sorour87 last updated on 04/Jan/18

$$L^{-1}\left(\frac{s^3}{s^4+4}\right)=?$$

Answered by sma3l2996 last updated on 04/Jan/18

$$\frac{s^3}{s^4+4}=\frac{a}{s-(1+i)}+\frac{b}{s+(1-i)}+\frac{c}{s+(1+i)}+\frac{d}{s-(1-i)}$$$$a=b=c=d=\frac{1}{4}$$$$\frac{s^3}{s^4+4}=\frac{1}{4}(\frac{1}{s-(1+i)}+\frac{1}{s+(1-i)}+\frac{1}{s+(1+i)}+\frac{1}{s-(1-i)})$$$$so:$$$$L^{-1}\left(\frac{s^3}{s^4+4}\right)=\frac{1}{4}(L^{-1}\left(\frac{1}{s-(1+i)}\right)+L^{-1}\left(\frac{1}{s+(1-i)}\right)+L^{-1}\left(\frac{1}{s+(1+i)}\right)+L^{-1}\left(\frac{1}{s-(1-i)}\right))$$$$=\frac{1}{4}(e^{(1+i)t}+e^{-(1-i)t}+e^{-(1+i)t}+e^{(1-i)t})$$$$=\frac{1}{4}(e^t\times e^{it}+e^{-t}\times e^{it}+e^{-t}\times e^{-it}+e^t\times e^{-it})$$$$=\frac{1}{4}(e^t+e^{-t})(e^{it}+e^{-it})$$$$so:L^{-1}\left(\frac{s^3}{s^4+4}\right)=cos\left(t\right)cosh\left(t\right)$$

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