Question

With initial conditions solve for y(t).
$y^{\prime \prime }\left(t\right)+5y^{\prime }\left(t\right)+6y\left(t\right)=x\left(t\right)$
$y\left(0^{-}\right)=2,y^{\prime }\left(0^{-}\right)=1andx\left(t\right)=e^{-t}u\left(t\right)$

• A. $\frac{1}{2}{e}^{-t}+6e^{-2t}+\frac{9}{2}{e}^{-3t}$
• B. $\frac{1}{2}{e}^{-t}-6e^{2t}+\frac{9}{2}{e}^{-3t}$
• C. $\frac{1}{2}{e}^{-t}+6e^{-2t}-\frac{9}{2}{e}^{-3t}$
• D. $\frac{1}{2}{e}^{-t}-6e^{-2t}-\frac{9}{2}{e}^{-3t}$

Answer 1

The correct option is C $\frac{1}{2}{e}^{-t}+6e^{-2t}-\frac{9}{2}{e}^{-3t}$

Taking Laplace transform on both side of the equation,

$\left[s^{2}Y\right(s)-2s-1]+5\left[sY\right(s)-2]+6Y\left(s\right)=\frac{1}{s+1}$

$\therefore Y\left(s\right)=\frac{2s^2+13s+12}{(s+1)(s^2+5s+6)}=\frac{2s^2+13s+12}{(s+1)(s+2)(s+3)}$

Using partial fraction expansion,

$Y\left(s\right)=\frac{1}{2(s+1)}+6.\frac{1}{(s+2)}-\frac{9}{2}.\frac{1}{(s+3)}$

$\therefore$ Taking the inverse Laplace transform of Y(s),

$y\left(t\right)=\frac{1}{2}{e}^{-t}+6e^{-2t}-\frac{9}{2}{e}^{-3t}$