# Solution to the boundary-value problem $-9 \frac{d^{2} u}{d x^{2}}+u=5 x, 0<x<3$ with $u(0)=0,\left.\frac{d u}{d x}\right|_{x=3}=0$ is (A) $u(x)=\frac{15 e}{1+e^{2}}\left(e^{-x / 3}-e^{x / 3}\right)+5 x$ (B) $u(x)=\frac{15 e}{1+e^{2}}\left(e^{-x / 3}+e^{x / 3}\right)+5 x$ (C) $u(x)=-\frac{15 \sin (x / 3)}{\cos (1)}+5 x$ (D) $u(x)=-\frac{15 \sin (x / 3)}{\cos (1)}-\frac{5}{54} x^{3}$

## Question ID - 155903 :- Solution to the boundary-value problem $-9 \frac{d^{2} u}{d x^{2}}+u=5 x, 0<x<3$ with $u(0)=0,\left.\frac{d u}{d x}\right|_{x=3}=0$ is (A) $u(x)=\frac{15 e}{1+e^{2}}\left(e^{-x / 3}-e^{x / 3}\right)+5 x$ (B) $u(x)=\frac{15 e}{1+e^{2}}\left(e^{-x / 3}+e^{x / 3}\right)+5 x$ (C) $u(x)=-\frac{15 \sin (x / 3)}{\cos (1)}+5 x$ (D) $u(x)=-\frac{15 \sin (x / 3)}{\cos (1)}-\frac{5}{54} x^{3}$

3537

(A) $u(x)=\frac{15 e}{1+e^{2}}\left(e^{-x / 3}-e^{x / 3}\right)+5 x$

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