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GATE 2016 EE – Question 14

Engineering Mathematics · Differential Equations: Initial and boundary value problems · 1 mark · Multiple choice

A function $y(t)$, such that $y(0) = 1$ and $y(1) = 3e^{-1}$, is a solution of the differential equation $\frac{d^2 y}{dt^2} + 2\frac{dy}{dt} + y = 0$. Then $y(2)$ is

  1. $5e^{-1}$
  2. $5e^{-2}$
  3. $7e^{-1}$
  4. $7e^{-2}$

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Correct answer: (B) $5e^{-2}$

Explanation

The characteristic equation $r^2 + 2r + 1 = 0$ has a repeated root $r = -1$, so $y = (A + Bt)e^{-t}$. From $y(0) = 1$ we get $A = 1$. From $y(1) = (1 + B)e^{-1} = 3e^{-1}$ we get $B = 2$. Then $y(2) = (1 + 4)e^{-2} = 5e^{-2}$.