Consider the initial value problem: $ \frac{d^{2} y}{d t^{2}}+4 \frac{d y}{d t}+6 y=f(t) ; y(0)=2,\left(\frac{d y}{d t}\right)_{t=0}=1$
If $Y(s)=\int_{0}^{\infty} y(t) e^{-s t} d t$ and $F(s)=\int_{0}^{\infty} f(t) e^{-s t} d t$ are the Laplace transforms of $y(t)$ and $f(t)$ respectively, then $Y(s)$ is given by:
$
\begin{array}{ll}
\text { (A) } \frac{F(s)}{\left(s^{2}+4 s+6\right)} & \text { (B) } \frac{F(s)+2 s+9}{\left(s^{2}+4 s+6\right)}\\
\text { (C) } \frac{F(s)}{\left(-s^{2}+4 s+6\right)} & \text { (D) } \frac{F(s)-2 s+9}{\left(s^{2}+4 s+6\right)}
\end{array}
$

Question ID - 155577 | SaraNextGen Top Answer

Consider the initial value problem: $ \frac{d^{2} y}{d t^{2}}+4 \frac{d y}{d t}+6 y=f(t) ; y(0)=2,\left(\frac{d y}{d t}\right)_{t=0}=1$
If $Y(s)=\int_{0}^{\infty} y(t) e^{-s t} d t$ and $F(s)=\int_{0}^{\infty} f(t) e^{-s t} d t$ are the Laplace transforms of $y(t)$ and $f(t)$ respectively, then $Y(s)$ is given by:
$
\begin{array}{ll}
\text { (A) } \frac{F(s)}{\left(s^{2}+4 s+6\right)} & \text { (B) } \frac{F(s)+2 s+9}{\left(s^{2}+4 s+6\right)}\\
\text { (C) } \frac{F(s)}{\left(-s^{2}+4 s+6\right)} & \text { (D) } \frac{F(s)-2 s+9}{\left(s^{2}+4 s+6\right)}
\end{array}
$