For the differential equationd2xdt2-6dxdt-8x-0 with initial conditions x0-1 and -dxdt-t-0-0- the solution is

D^2xdt^2+6dxdt+8x=0 A.E is D^2+6D+8=0 D= 2, 4 x=C_1e^ 2t+C_2e^ 4t dxdt= 2C_1e^ 2t 4C_2e^ 4t x(0)=1 ⇒ C_1+C_2=1 | dxdt |_t=0=0 ⇒ C_1+2C_2=0 So, nbsp; ...

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For the diffe...

Question

For the differential equation
$\frac{d^{2} x}{d t^{2}} + 6 \frac{d x}{d t} + 8 x = 0$ with initial conditions $x (0) = 1$ and ${\begin{matrix} \frac{d x}{d t} \end{matrix} ∣ ∣ ∣}_{t = 0} = 0)$ , the solution is

x (t) = 2 e^{- 6 t} - e^{- 2 t}

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x (t) = 2 e^{- 2 t} - e^{- 4 t}

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x (t) = - e^{- 6 t} + 2 e^{- 4 t}

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a (t) = e^{- 2 t} + 2 e^{- 4 t}

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Solution

The correct option is B $x (t) = 2 e^{- 2 t} - e^{- 4 t}$
$\frac{d^{2} x}{d t^{2}} + 6 \frac{d x}{d t} + 8 x = 0$
$A . E$ is $D^{2} + 6 D + 8 = 0$
$D = - 2, - 4$
$x = C_{1} e^{- 2 t} + C_{2} e^{- 4 t}$
$\frac{d x}{d t} = - 2 C_{1} e^{- 2 t} - 4 C_{2} e^{- 4 t}$
$x (0) = 1 \Rightarrow C_{1} + C_{2} = 1$
${∣ ∣ ∣ \frac{d x}{d t} ∣ ∣ ∣}_{t = 0} = 0$
$\Rightarrow C_{1} + 2 C_{2} = 0$
So, $C_{1} = 2, C_{2} = - 1$
$x (t) = 2 e^{- 2 t} - e^{- 4 t}$

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For the differential equation d2xdt2+6dxdt+8x=0 with initial conditions x(0)=1 and dxdt∣∣∣t=0=0), the solution is

The correct option is B x(t)=2e−2t−e−4td2xdt2+6dxdt+8x=0 A.E is D2+6D+8=0 D=−2,−4 x=C1e−2t+C2e−4t dxdt=−2C1e−2t−4C2e−4t x(0)=1⇒C1+C2=1 ∣∣∣dxdt∣∣∣t=0=0 ⇒C1+2C2=0 So, C1=2,C2=−1 x(t)=2e−2t−e−4t

For the differential equation
$\frac{d^{2} x}{d t^{2}} + 6 \frac{d x}{d t} + 8 x = 0$ with initial conditions $x (0) = 1$ and ${\begin{matrix} \frac{d x}{d t} \end{matrix} ∣ ∣ ∣}_{t = 0} = 0)$ , the solution is