MDoubt No 000016 - GATE

GATE [EE, 2010, 2 marks]
Question : For the differential equation $\displaystyle\frac{d^2x}{dt^2}+6\frac{dx}{dt}+8x=0$ with initial conditions $x(0)=1$ and $\displaystyle\frac{dx}{dt}\Bigg|_{t=0}=0$, the solution is

(a) $x(t)=2e^{-6t}-e^{-2t}$
(b) $x(t)=2e^{-2t}-e^{-4t}$
(c) $x(t)=-e^{-6t}+2e^{-4t}$
(d) $x(t)=e^{-2t}+2e^{-4t}$

Solution : Given $\dfrac{d^2x}{dt^2}+6\dfrac{dx}{dt}+8x=0$
Let $\dfrac{d}{dt}=D$
so the equation becomes
$\implies (D^2+6D+8)x=0$
the auxiliary equation is
$\implies D^2+6D+8=0$
$\implies (D+2)(D+4)=0$
$\implies D=-2,-4$
the solution is
$\implies x=c_1e^{-2t}+c_2e^{-4t}\quad\quad\quad\quad\ldots(1)$
differentiting equation $(1)$
$\implies\dfrac{dx}{dt}=-2c_1e^{-2t}-4c_2e^{-4t}\quad\quad\quad\quad\ldots(2)$
since $x(0)=1$ and $\dfrac{dx}{dt}\Bigg|_{t=0}=0$
putting values in equations $(1)$ and $(2)$ respectively
$1=c_1+c_2\quad\quad\quad\quad\ldots(3)$
$0=-2c_1-4c_2\quad\quad\quad\quad\ldots(4)$
solving $3$ and $(4)$
$\implies c_1=2, c_2=-1$
$\therefore$ the solution is
$x(t)=2e^{-2t}-e^{-4t}$
CORRECT ANSWER : B