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The Laplace transform $L(\mathrm{u}(\mathrm{t}))=\mathrm{U}(\mathrm{s}),$ for the solution $\mathrm{u}(\mathrm{t})$ of the problem $\frac{d^{2} u}{d t^{2}}+2 \frac{d u}{d t}+u=1, t>0$
with initial conditions $u(0)=0, \frac{d u(0)}{d t}=5$ is given by:
(A) $\frac{6}{(s+1)^{2}}$
(B) $\frac{5 s+1}{s(s+1)^{2}}$
(C) $\frac{1-5 s}{s(s+1)^{2}}$
(D) $\frac{5 s^{2}+1}{s(s+1)^{2}}$



Question ID - 155904 | SaraNextGen Top Answer

The Laplace transform $L(\mathrm{u}(\mathrm{t}))=\mathrm{U}(\mathrm{s}),$ for the solution $\mathrm{u}(\mathrm{t})$ of the problem $\frac{d^{2} u}{d t^{2}}+2 \frac{d u}{d t}+u=1, t>0$
with initial conditions $u(0)=0, \frac{d u(0)}{d t}=5$ is given by:
(A) $\frac{6}{(s+1)^{2}}$
(B) $\frac{5 s+1}{s(s+1)^{2}}$
(C) $\frac{1-5 s}{s(s+1)^{2}}$
(D) $\frac{5 s^{2}+1}{s(s+1)^{2}}$

1 Answer
127 votes
Answer Key / Explanation : (B) -

(B) $\frac{5 s+1}{s(s+1)^{2}}$

127 votes


127