# 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 :- 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) $\frac{5 s^{2}+1}{s(s+1)^{2}}$

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

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For a steady, incompressible two-dimensional flow, represented in Cartesian co-ordinates $(x, y),$ a student correctly writes the equation of pathline of any arbitrary particle as, $\frac{d x}{d t}=a x$ and $\frac{d y}{d t}=b y$, where $a$ and bare constants having unit of (second) $^{-1}$. If value of $a$ is $5,$ the value of $b$ is_____________ 