Match the third order linear equations with their fundamental solution sets. 1. y′′′−6y′′+y′−6y=0 2. y′′′−8y′′+15y′=0 3. y′′′+y′=0 4. y′′′−y

Question

Match the third order linear equations with their fundamental solution sets. 1. y′′′−6y′′+y′−6y=0 2. y′′′−8y′′+15y′=0 3. y′′′+y′=0 4. y′′′−y′′−y′+y=0 5. ty′′′−y′′=0 6. y′′′+3y′′+3y′+y=0

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Ximena 4 days 2021-10-11T13:57:40+00:00 1 Answer 0

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    2021-10-11T13:58:57+00:00

    The fundamental solution sets to the third order linear equations are not given. I would however, find the solutions, so that you can match them to their corresponding options.

    Answer:

    (1) y”’ – 6y” + y’ – 6y = 0

    Corresponds to:

    y = C1e^(6t) + C2cost + C3sint

    (2) y”’ – 8y” + 15y’ = 0

    Corresponds to:

    y = C1 + C2e^(3t) + C3e^(5t)

    (3) y”’ + y’ = 0

    Corresponds to:

    y = C1 + C2cost + C3sint

    (4) y”’ – y” – y’ + y = 0

    Corresponds to:

    y = C1e^(-t) + (C2 + C3t) e^t

    (5) ty”’ – y” = 0

    Corresponds to:

    y = C1 + C2t + C3t³

    (6) y”’ + 3y” + 3y’ + y = 0

    Corresponds to:

    y = (C1 + C2t + C3t²)e^(-t)

    Step-by-step explanation:

    Given the following differential equations

    (1) y”’ – 6y” + y’ – 6y = 0

    (2) y”’ – 8y” + 15y’ = 0

    (3) y”’ + y’ = 0

    (4) y”’ – y” – y’ + y = 0

    (5) ty”’ – y” = 0

    (6) y”’ + 3y” + 3y’ + y = 0

    SOLUTIONS

    (1) y”’ – 6y” + y’ – 6y = 0

    We write the characteristic equation and solve.

    The characteristic equation is

    m³ – 6m² + m – 6 = 0

    m²(m – 6) + (m – 6) = 0

    (m² + 1)(m – 6) = 0

    m² + 1 = 0

    => m² = -1

    => m = ±√(-1) = ±i

    Or m – 6 = 0

    => m = 6

    The solutions are therefore

    m = ±i, 6

    Therefore, the solution to the differential equation is

    y = C1e^(6t) + C2cost + C3sint

    (2) y”’ – 8y” + 15y’ = 0

    The characteristic equation is

    m³ – 8m² + 15m = 0

    m³ – 5m² – 3m² + 15m = 0

    m²(m – 5) – 3m(m – 5) = 0

    (m² – 3m)(m – 5) = 0

    m(m – 3)(m – 5) = 0

    m = 0

    Or

    m – 3 = 0

    => m = 3

    Or

    m – 5 = 0

    => m = 5

    Therefore,

    y = C1 + C2e^(3t) + C3e^(5t)

    (3) y”’ + y’ = 0

    The characteristic equation is

    m³ + m = 0

    m(m² + 1) = 0

    m = 0

    Or

    m² + 1 = 0

    => m² = -1

    => m = ±√(-1) = ±i

    Therefore,

    y = C1 + C2cost + C3sint

    (4) y”’ – y” – y’ + y = 0

    The characteristic equation is

    m³ – m² – m + 1 = 0

    m²(m – 1) – (m – 1) = 0

    (m² – 1)(m – 1) = 0

    (m – 1)(m + 1)(m – 1) = 0

    Then

    m – 1 = 0 twice

    => m = 1 twice

    Or

    m + 1 = 0

    => m = -1

    Therefore,

    y = C1e^(-t) + (C2 + C3t) e^t

    (5) ty”’ – y” = 0

    Multiplying by t², this can be written as

    t³y”’ – t²y” = 0

    Put t = e^z, so that z = lnt, and Dy = tdy/dt

    Then

    dy/dt = (1/t)dy/dz

    ty’ = dy/dz

    Again,

    d²y/dt² = -(1/t²)dy/dz + (1/t)(d²y/dz²)(dz/dt)

    => t²y” = D(D – 1)y

    Similarly,

    t³y”’ = D(D – 1)(D – 2)y

    Using these, we have

    [D(D – 1)(D – 2) – D(D – 1)]y = 0

    D(D – 1)(D – 3)y = 0

    The characteristic equation is

    m(m – 1)(m – 3) = 0

    m = 0

    Or

    m – 1 = 0

    => m = 1

    Or

    m – 3 = 0

    => m = 3

    y = C1 + C2e^z + C3(e^z)³

    But e^z = t

    So

    y = C1 + C2t + C3t³

    (6) y”’ + 3y” + 3y’ + y = 0

    The characteristic equation is

    m³ + 3m² + 3m + 1 = 0

    (m + 1)³ = 0

    m = -1 three times

    Therefore,

    y = (C1 + C2t + C3t²)e^(-t)

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