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The solution of the differential equation, for t > 0, y″(t) + 2y′(t) + y(t) = 0 with initial conditions y′(0) = 1and y(0) = 0, is u(t) denotes the unit step functions).
  • a)
    te -t u(t)
  • b)
    (e-t - te-t) u (t)
  • c)
    (-e-t + te-t u (t))
  • d)
    e-t u (t)
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The solution of the differential equation, fort > 0,y″(t) + 2...
Concept:
Laplace transform of the first two derivatives.
L{y''} = s2Y(s) - sy(0) - y'(0)
L{y'} = sY(s) - y(0)
Calculation:
Given y″(t) + 2y′(t) + y(t) = 0
Apply Laplace transform we get
[s2Y(s) − sy(0) − y′(0)] + 2[sy(s) − y(0)] + Y(s) = 0
(s2 + 2s + 1)Y(s)
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The solution of the differential equation, fort > 0,y″(t) + 2y′(t) + y(t) = 0with initial conditionsy′(0) = 1and y(0) = 0, isu(t)denotes the unit step functions).a)te-tu(t)b)(e-t - te-t) u (t)c)(-e-t + te-t u (t))d)e-t u (t)Correct answer is option 'A'. Can you explain this answer?
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