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17 October, 23:54

Suppose that r1 and r2 are roots of ar2 + br + c = 0 and that r1 = r2; then exp (r1t) and exp (r2t) are solutions of the differential equation ay + by + cy = 0. Show that φ (t; r1, r2) = er2t - er1t r2 -

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  1. 18 October, 03:39
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    The Correct Question is:

    Suppose that r1 and r2 are roots of ar² + br + c = 0 and that r1 = r2; then e^ (r1t) and e^ (r2t) are solutions of the differential equation

    ay'' + by' + cy = 0.

    Show that

    φ (t; r1, r2) = [e^ (r2t) - e^ (r1t) ] / (r2 - r1)

    is a solution of the differential equation.

    Answer:

    φ (t; r1, r2) is a solution of the differential equation, and it shown.

    Step-by-step explanation:

    Given the differential equation

    ay'' + by' + cy = 0

    and r1 and r2 are the roots of its auxiliary equation.

    We want to show that

    φ (t; r1, r2) = [e^ (r2t) - e^ (r1t) ] / (r2 - r1)

    satisfies the given differential equation, that is

    aφ'' + bφ' + cφ = 0 ... (*)

    Where φ = φ (t; r1, r2)

    We now differentiate φ twice in succession, with respect to t.

    φ' = [r2e^ (r2t) - r1e^ (r1t) ] / (r2 - r1)

    φ'' = [r2²e^ (r2t) - r1²e^ (r1t) ] / (r2 - r1)

    Using these in (*)

    We have

    a[r2e^ (r2t) - r1e^ (r1t) ] / (r2 - r1) + [r2²e^ (r2t) - r1²e^ (r1t) ] / (r2 - r1) + c[e^ (r2t) - e^ (r1t) ] / (r2 - r1)

    = [ (ar2² + br2 + c) e^ (r2t) - (ar1² + br1 + c) e^ (r1t) ] / (r1 - r2)

    We know that r1 and r2 are the roots of the auxiliary equation

    ar² + br + c = 0

    and r1 = r2

    This implies that

    ar1² + br1 + c = ar2² + br2 + c = 0

    And hence,

    [ (ar2² + br2 + c) e^ (r2t) - (ar1² + br1 + c) e^ (r1t) ] / (r1 - r2) = 0

    Therefore,

    aφ'' + bφ' + cφ = 0
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