Consider the differential equation $$\frac{{{d^2}x}}{{d{t^2}}} + 2\frac{{dx}}{{dt}} + x = 0$$
At time t = 0, it is given that x = 1 and $$\frac{{dx}}{{dt}} = 0.$$ At t = 1, the value of x is given by
A. $$\frac{1}{e}$$
B. $$\frac{2}{e}$$
C. 1
D. $$\frac{3}{e}$$
Answer: Option B


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