Question

For the differential equation \(\frac{{{d^2}x}}{{d{t^2}}} + 6\frac{{dx}}{{dt}} + 8x = 0\) with initial conditions x(0)=1 and \({\left. {\frac{{dx}}{{dt}}} \right|_{t = 0}} = 0,\) the solution is
1. x(t) = 2e ^-6t - e^-2t
2. x(t) = 2e^{- 2t} - e^-4t
3. x(t) = -e ^-6t + 2e ^-4t
4. x(t) = e^-2t  + 2e ^-4t

Answer 1

Correct Answer - Option 2 : x(t) = 2e^{- 2t} - e^-4t

\(\frac{{{d^2}x}}{{d{t^2}}} + \frac{{6dx}}{{dt}} + 8x = 0\)

Taking laplace transform (with initial condition) on both sides

S² X(s) – sx(o) –x^’(o) + 6 [sX (s) – x (o) ] +8X(s) =0

S²X (s) – s(1)  - 0 + 6 [s X(s) -1] +8X(s) =0

X(s) [s²+6s+8] –s-6=0

\(X\left( s \right) = \frac{{s + 6}}{{{s^2} + 6s + 8}}\)

By partial fraction,

\(X\left( s \right) = \frac{2}{{s + 2}} - \frac{1}{{s + 4}}\)

Taking inverse laplace transform,

x(t) = (2e^-2t – e^-4t)