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The Laplace transform of eatcosωt e^{at} cos ωteatcosωt​
Question

The Laplace transform of eatcosωt e^{at} cos ωt

A.

(sa)(sa)2+ω2\quad \frac{(s - a)}{(s - a)^2 + \omega^2} \\

B.

ω(sa)2+ω2\quad \frac{\omega}{(s - a)^2 + \omega^2} \\

C.

a(sa)2+ω2\quad \frac{a}{(s - a)^2 + \omega^2} \\

D.

S(sa)2+ω2\quad \frac{S}{(s - a)^2 + \omega^2}

Correct option is A

The Laplace transform of eatcos(ωt) is given by:L{eatcos(ωt)}=sa(sa)2+ω2Derivation:1. Standard Laplace Transform of cos(ωt):L{cos(ωt)}=ss2+ω22. Applying the Exponential Shift Theorem:If L{f(t)}=F(s), then:L{eatf(t)}=F(sa)Applying this to cos(ωt):L{eatcos(ωt)}=sa(sa)2+ω2\text{The Laplace transform of } e^{at} \cos(\omega t) \text{ is given by:} \\[6pt]\mathcal{L}\{e^{at} \cos(\omega t)\} = \frac{s - a}{(s - a)^2 + \omega^2} \\[12pt]\textbf{Derivation:} \\[6pt]1. \ \textbf{Standard Laplace Transform of } \cos(\omega t): \\[4pt]\mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2 + \omega^2} \\[10pt]2. \ \textbf{Applying the Exponential Shift Theorem:} \\[4pt]\text{If } \mathcal{L}\{f(t)\} = F(s), \text{ then:} \\[4pt]\mathcal{L}\{e^{at} f(t)\} = F(s - a) \\[10pt]\text{Applying this to } \cos(\omega t): \\[4pt]\mathcal{L}\{e^{at} \cos(\omega t)\} = \frac{s - a}{(s - a)^2 + \omega^2}​​

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