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15 March, 22:11

In this problem, y = c1ex + c2e-x is a two-parameter family of solutions of the second-order DE y'' - y = 0. Find c1 and c2 given the following initial conditions. (Your answers will not contain a variable.) y (1) = 0, y' (1) = e c1 = Incorrect: Your answer is incorrect. c2 = Incorrect: Your answer is incorrect. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions. y = Incorrect: Your answer is incorrect.

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  1. 16 March, 00:02
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    c₁ = 1/2

    c₂ = - e²/2

    y = (1/2) * (eˣ - e²⁻ˣ)

    Step-by-step explanation:

    Given

    y = c₁eˣ + c₂e⁻ˣ

    y (1) = 0

    y' (1) = e

    We get y':

    y' = (c₁eˣ + c₂e⁻ˣ) ' ⇒ y' = c₁eˣ - c₂e⁻ˣ

    then we find y (1):

    y (1) = c₁e¹ + c₂e⁻¹ = 0

    ⇒ c₁ = - c₂/e² (I)

    then we obtain y' (1):

    y' (1) = c₁e¹ - c₂e⁻¹ = e (II)

    ⇒ ( - c₂/e²) * e - c₂e⁻¹ = e

    ⇒ - c₂e⁻¹ - c₂e⁻¹ = - 2c₂e⁻¹ = e

    ⇒ c₂ = - e²/2

    and

    c₁ = - c₂/e² = - ( - e²/2) / e²

    ⇒ c₁ = 1/2

    Finally, the equation will be

    y = (1/2) * eˣ - (e²/2) * e⁻ˣ = (1/2) * (eˣ - e²⁻ˣ)
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