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Homogeneous Linear Differential Equations (General Form of LDE)

Consider the ...

Question

Consider the differential equation $. . y + 2 . y + y = 0$ with boundary conditions $y (0) = 1$ & $y (1) = 0$ . The value of $y (2)$ is

- 1

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- e^{- 1}

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- e^{- 2}

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e^{2}

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Solution

The correct option is C $- e^{- 2}$
Given, $(D^{2} + 2 D + 1) y = 0$ ....(1)
A.E. is $m^{2} + 2 m + 1 = 0$
$m = - 1, - 1$
So, $y = (C_{1} + C_{2} x) e^{- x}$ ....(2)
Using $y (0) = 1, C_{1} = 1$
Using $y (1) = 0, C_{2} = - 1$
So, solution of $(1)$ is
$y = (1 - x) e^{- x}$
So, $y (2) = (1 - 2) e^{- 2} = \frac{- 1}{e^{2}}$

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