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28 March, 07:53

f1 (x) = ex, f2 (x) = e-x, f3 (x) = sinh (x) g (x) = c1f1 (x) + c2f2 (x) + c3f3 (x) Solve for c1, c2, and c3 so that g (x) = 0 on the interval (-[infinity], [infinity]). If a nontrivial solution exists, state it. (If only the trivial solution exists, enter the trivial solution {0, 0, 0}.)

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  1. 28 March, 08:04
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    (C1, C2, C3) = (K, K, - 2K)

    For K in the interval (-∞, ∞)

    Step-by-step explanation:

    Given

    f1 (x) = e^x

    f2 (x) = e^ (-x)

    f3 (x) = sinh (x)

    g (x) = 0

    We want to solve for C1, C2 and C3, such that

    C1f1 (x) + C2f2 (x) + C3f3 (x) = g (x)

    That is

    C1e^x + C2e^ (-x) + C3sinh (x) = 0

    The hyperbolic sine of x, sinh (x), can be written in its exponential form as

    sinh (x) = (1/2) (e^x + e^ (-x))

    So, we can rewrite

    C1e^x + C2e^ (-x) + C3sinh (x) = 0

    as

    C1e^x + C2e^ (-x) + C3 (1/2) (e^x + e^ (-x)) = 0

    So we have

    (C1 + (1/2) C3) e^x + (C2 + (1/2) C3) e^ (-x) = 0

    We know that

    e^x ≠ 0, and e^ (-x) ≠ 0

    So we must have

    (C1 + (1/2) C3) = 0 ... (1)

    and

    (C2 + (1/2) C3) = 0 ... (2)

    From (1)

    2C1 + C3 = 0

    => C3 = - 2C1 ... (3)

    From (2)

    2C2 + C3 = 0

    => C3 = - 2C2 ... (4)

    Comparing (3) and (4)

    2C1 = 2C2

    => C2 = C1

    Let C1 = C2 = K

    C3 = - 2K

    (C1, C2, C3) = (K, K, - 2K)

    For K in the interval (-∞, ∞)
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