Question #54517

pls solve inerse laplace problem :

L^-1{ a/(s+b)*exp(-y *sqrt(s+c))} .

Expert's answer

Answer on Question #54517 – Math – Differential Equations

pls solve inverse laplace problem :


L1{a/(s+b)exp(ysqrt(s+c))}.L ^ {\wedge} - 1 \left\{a / (s + b) ^ {*} \exp (- y ^ {*} s q r t (s + c)) \right\}.


Solution


f(t)=L1{as+beys+c}f (t) = L ^ {- 1} \left\{\frac {a}{s + b} e ^ {- y \sqrt {s + c}} \right\}L1{aseys+c}=a2eycerfc(y2ct2t)+a2eycerfc(y+2ct2t), y>0.L ^ {- 1} \left\{\frac {a}{s} e ^ {- y \sqrt {s + c}} \right\} = \frac {a}{2} e ^ {- y \sqrt {c}} \operatorname {erfc} \left(\frac {y - 2 \sqrt {c t}}{2 \sqrt {t}}\right) + \frac {a}{2} e ^ {y \sqrt {c}} \operatorname {erfc} \left(\frac {y + 2 \sqrt {c t}}{2 \sqrt {t}}\right), \ y > 0.


If f(t)=L1{F(s)}f(t) = L^{-1}\{F(s)\}, then f(t)=eatL1{F(s+a)}f(t) = e^{at}L^{-1}\{F(s + a)\}.

Thus f(t)=ebtL1{aseys+cb}=f(t) = e^{bt}L^{-1}\left\{\frac{a}{s} e^{-y\sqrt{s + c - b}}\right\} =

=a2ebt[eycberfc(y2cbt2t)+eycberfc(y+2cbt2t)]= \frac {a}{2} e ^ {b t} \left[ e ^ {- y \sqrt {c - b}} \operatorname {erfc} \left(\frac {y - 2 \sqrt {c - b} t}{2 \sqrt {t}}\right) + e ^ {y \sqrt {c - b}} \operatorname {erfc} \left(\frac {y + 2 \sqrt {c - b} t}{2 \sqrt {t}}\right) \right]


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