Question

laplace transform of s/(s^2+a^2)^2​

Answer 1

❑ CORRECT QUESTION :-

$\sf \large Find \: th e \: laplace \: \: inverse \: of \: -$

$\sf \: \: \frac{s}{( { {s}^{2} + {a}^{2}) }^{2} }$

❑ SOLUTION :-

❑ GIVEN :-

$\sf \: F(s) \: \: or \: \: L \{ f(t) \} = \: \frac{s}{( { {s}^{2} + {a}^{2}) }^{2} }$

$\sf \: We \: \: know \: \: the \: \: formula -$

$\sf \: L\{t.f(t)\} = - \frac{d}{ds} (f(s))$

$\sf \: Then$

$\sf \: t.f(t)= {L}^{ - 1} \bigg( - \frac{d}{ds} (f(s)) \bigg)$

$\sf \: Or \: \: we \: \: can \: \: say \: \: that -$

$\orange{ \sf \: {L}^{ - 1} \bigg( \frac{d}{ds} (f(s)) \bigg) = - \: t.f(t)}$

$\sf \: From \: \: the \: \: Question$

$\sf \: Let$

$\sf \: f(s) \: = \frac{1}{( { {s}^{2} + {a}^{2}) } }$

$\sf \: Then$

$\sf \: f(t) \: = L(f(s)) = L \bigg( \frac{1}{ ({ {s}^{2} + {a}^{2} ) } } \bigg)$

$\sf \: f(t) = \: \: \frac{ \sin(at) }{a}$

$\sf \: Now \: \: appling \: \: the \: \: formula -$

$\sf \: {L}^{ - 1} \bigg( \frac{d}{ds} (f(s)) \bigg) = - \: t.f(t)$

$\implies\sf \: {L}^{ - 1} \bigg( \frac{d}{ds} \bigg \{ \frac{1}{ {s}^{2} + {a}^{2} } \bigg \} \bigg) = - \: t. \: \frac{ \sin(at) }{a}$

$\implies\sf \: {L}^{ - 1} \bigg( \frac{ - 1}{ ({s}^{2} + {a}^{2}) ^{2} } (2s) \bigg) = - \: t. \: \frac{ \sin(at) }{a}$

$\implies\sf \: - {L}^{ - 1} \bigg( \frac{ 2s}{ ({s}^{2} + {a}^{2}) ^{2} } \bigg) = - \: t. \: \frac{ \sin(at) }{a}$

$\implies\sf \: {L}^{ - 1} \bigg( \frac{ 2s}{ ({s}^{2} + {a}^{2}) ^{2} } \bigg) = \: t. \: \frac{ \sin(at) }{a}$

$\sf \: Now \: \: divided \: \: by \: \: 2 \: \: in \: \: both \: \: sides -$

$\implies\sf \: {L}^{ - 1} \bigg( \frac{ s}{ ({s}^{2} + {a}^{2}) ^{2} } \bigg) = \: t. \: \frac{ \sin(at) }{2a}$

$\sf \: Hence \: Required \: Inverse \: \: Laplace \: is </p><p>\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: t. \: \frac{ \sin(at) }{2a}$