MA 303 Homework 1 (Homogeneous Linear Differential MA 303 Homework 1 (Homogeneous Linear...

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  • MA 303 Homework 1

    (Homogeneous Linear Differential Equations)

    Hoon Hong

    1st-order, 1 variable Problem:

    y'K2 y = 0 y(0) =K3

    1. Find the general solution: λK2 = 0

    λ = 2

    y = C e2 t

    2. Find the particular solution: C =K3

    y =K3 e2 t

    3. Sketch the particular solution:

    t

    y

    t

    1 of 48

  • Problem: y'K2 y = 0 y(0) = 4

    1. Find the general solution: λK2 = 0

    λ = 2

    y = C e2 t

    2. Find the particular solution: C = 4

    y = 4 e2 t

    3. Sketch the particular solution:

    t

    y

    t

    2 of 48

  • Problem: y'C3 y = 0 y(0) = 2

    1. Find the general solution: λC3 = 0

    λ =K3

    y = C eK3 t

    2. Find the particular solution: C = 2

    y = 2 eK3 t

    3. Sketch the particular solution:

    t

    y

    t

    3 of 48

  • Problem: y'C3 y = 0 y(0) =K2

    1. Find the general solution: λC3 = 0

    λ =K3

    y = C eK3 t

    2. Find the particular solution: C =K2

    y =K2 eK3 t

    3. Sketch the particular solution:

    t

    y

    t

    4 of 48

  • 2nd-order, 1 variable: Real eigenvalues Problem:

    y''C3 y'C2 y = 0 y(0) =K3 y'(0) = 2

    1. Find the general solution: λ

    2 C3 λC2 = 0

    λ =K1, K2

    y = C1 e KtCC2 e

    K2 t

    2. Find the particular solution: y' =KC1 e

    KtK2 C2 e K2 t

    C1CC2 =K3

    KC1K2 C2 = 2

    C1 =K4, C2 = 1

    y =K4 eKtCeK2 t

    3. Sketch the particular solution:

    t

    y

    t

    5 of 48

  • Problem: y''K3 y'C2 y = 0 y(0) = 3 y'(0) = 1

    1. Find the general solution: λ

    2 K3 λC2 = 0

    λ = 2, 1

    y = C1 e 2 tCC2 e

    t

    2. Find the particular solution: y' = 2 C1 e

    2 tCC2 e t

    C1CC2 = 3

    2 C1CC2 = 1

    C1 =K2, C2 = 5

    y =K2 e2 tC5 et

    3. Sketch the particular solution:

    t

    y

    t

    6 of 48

  • Problem: y''Cy'K6 y = 0 y(0) =K3 y'(0) = 4

    1. Find the general solution: λ

    2 CλK6 = 0

    λ = 2, K3

    y = C1 e 2 tCC2 e

    K3 t

    2. Find the particular solution: y' = 2 C1 e

    2 tK3 C2 e K3 t

    C1CC2 =K3

    2 C1K3 C2 = 4

    C1 =K1, C2 =K2

    y =Ke2 tK2 eK3 t

    3. Sketch the particular solution:

    t

    y

    t

    7 of 48

  • Problem: y''C5 y'C6 y = 0 y(0) = 3 y'(0) =K8

    1. Find the general solution: λ

    2 C5 λC6 = 0

    λ =K2, K3

    y = C1 e K2 tCC2 e

    K3 t

    2. Find the particular solution: y' =K2 C1 e

    K2 tK3 C2 e K3 t

    C1CC2 = 3

    K2 C1K3 C2 =K8

    C1 = 1, C2 = 2

    y = eK2 tC2 eK3 t

    3. Sketch the particular solution:

    t

    y

    t

    8 of 48

  • Problem: y''C7 y'C12 y = 0 y(0) = 6 y'(0) =K22

    1. Find the general solution: λ

    2 C7 λC12 = 0

    λ =K3, K4

    y = C1 e K3 tCC2 e

    K4 t

    2. Find the particular solution: y' =K3 C1 e

    K3 tK4 C2 e K4 t

    C1CC2 = 6

    K3 C1K4 C2 =K22

    C1 = 2, C2 = 4

    y = 2 eK3 tC4 eK4 t

    3. Sketch the particular solution:

    t

    y

    t

    9 of 48

  • Problem: y''K5 y'C6 y = 0 y(0) = 2 y'(0) = 1

    1. Find the general solution: λ

    2 K5 λC6 = 0

    λ = 3, 2

    y = C1 e 3 tCC2 e

    2 t

    2. Find the particular solution: y' = 3 C1 e

    3 tC2 C2 e 2 t

    C1CC2 = 2

    3 C1C2 C2 = 1

    C1 =K3, C2 = 5

    y =K3 e3 tC5 e2 t

    3. Sketch the particular solution:

    t

    y

    t

    10 of 48

  • Problem: y''Cy'K2 y = 0 y(0) =K4 y'(0) = 5

    1. Find the general solution: λ

    2 CλK2 = 0

    λ = 1, K2

    y = C1 e tCC2 e

    K2 t

    2. Find the particular solution: y' = C1 e

    tK2 C2 e K2 t

    C1CC2 =K4

    C1K2 C2 = 5

    C1 =K1, C2 =K3

    y =KetK3 eK2 t

    3. Sketch the particular solution:

    t

    y

    t

    11 of 48

  • Problem: y''C5 y'C6 y = 0 y(0) =K1 y'(0) = 1

    1. Find the general solution: λ

    2 C5 λC6 = 0

    λ =K2, K3

    y = C1 e K2 tCC2 e

    K3 t

    2. Find the particular solution: y' =K2 C1 e

    K2 tK3 C2 e K3 t

    C1CC2 =K1

    K2 C1K3 C2 = 1

    C1 =K2, C2 = 1

    y =K2 eK2 tCeK3 t

    3. Sketch the particular solution:

    t

    y

    t

    12 of 48

  • 2nd-order, 1 variable: Non-real eigenvalues Problem:

    y''K4 y'C68 y = 0 y(0) = 1 y'(0) = 10

    1. Find the general solution: λ

    2 K4 λC68 = 0

    λ = 2C8 I, 2K8 I

    y = e2 t C1 cos 8 t CC2 sin 8 t

    2. Find the particular solution: y' = 2 e2 t C1 cos 8 t CC2 sin 8 t Ce

    2 t K8 C1 sin 8 t C8 C2 cos 8 t

    C1 = 1

    2 C1C8 C2 = 10

    C2 = 1

    y = e2 t cos 8 t Csin 8 t 3. Sketch the particular solution:

    t

    y

    t

    13 of 48

  • Problem: y''C2 y'C50 y = 0 y(0) = 2 y'(0) = 19

    1. Find the general solution: λ

    2 C2 λC50 = 0

    λ =K1C7 I, K1K7 I

    y = eKt C1 cos 7 t CC2 sin 7 t

    2. Find the particular solution: y' =KeKt C1 cos 7 t CC2 sin 7 t Ce

    Kt K7 C1 sin 7 t C7 C2 cos 7 t

    C1 = 2

    KC1C7 C2 = 19

    C2 = 3

    y = eKt 2 cos 7 t C3 sin 7 t 3. Sketch the particular solution:

    t

    y

    t

    14 of 48

  • Problem: y''C64 y = 0 y(0) = 2 y'(0) = 8

    1. Find the general solution: λ

    2 C64 = 0

    λ = 8 I, K8 I

    y = C1 cos 8 t CC2 sin 8 t

    2. Find the particular solution: y' =K8 C1 sin 8 t C8 C2 cos 8 t

    C1 = 2

    8 C2 = 8

    C2 = 1

    y = 2 cos 8 t Csin 8 t 3. Sketch the particular solution:

    t

    y

    t

    15 of 48

  • Problem: y''C2 y'C82 y = 0 y(0) = 0 y'(0) = 9

    1. Find the general solution: λ

    2 C2 λC82 = 0

    λ =K1C9 I, K1K9 I

    y = eKt C1 cos 9 t CC2 sin 9 t

    2. Find the particular solution: y' =KeKt C1 cos 9 t CC2 sin 9 t Ce

    Kt K9 C1 sin 9 t C9 C2 cos 9 t

    C1 = 0

    KC1C9 C2 = 9

    C2 = 1

    y = eKt sin 9 t 3. Sketch the particular solution:

    t

    y

    t

    16 of 48

  • Problem: y''K4 y'C125 y = 0 y(0) = 1 y'(0) = 35

    1. Find the general solution: λ

    2 K4 λC125 = 0

    λ = 2C11 I, 2K11 I

    y = e2 t C1 cos 11 t CC2 sin 11 t

    2. Find the particular solution: y' = 2 e2 t C1 cos 11 t CC2 sin 11 t Ce

    2 t K11 C1 sin 11 t C11 C2 cos 11 t

    C1 = 1

    2 C1C11 C2 = 35

    C2 = 3

    y = e2 t cos 11 t C3 sin 11 t 3. Sketch the particular solution:

    t

    y

    t

    17 of 48

  • Problem: y''C9 y = 0 y(0) = 1 y'(0) =K3

    1. Find the general solution: λ

    2 C9 = 0

    λ = 3 I, K3 I

    y = C1 cos 3 t CC2 sin 3 t

    2. Find the particular solution: y' =K3 C1 sin 3 t C3 C2 cos 3 t

    C1 = 1

    3 C2 =K3

    C2 =K1

    y = cos 3 t Ksin 3 t 3. Sketch the particular solution:

    t

    y

    t

    18 of 48

  • 1st-order, 2 variables: Introduction Problem:

    y1' =Ky1C2 y2 y2' = 2 y1C2 y2 y1(0) = 3 y2(0) = 1

    1. Find the general solution: K1Kλ v1C2 v2 = 0

    2 v1C 2Kλ v2 = 0

    λ 2 KλK6 = 0

    λ = 3

    v1 = 2 C1 v2 = 4 C1

    λ =K2

    v1 = 2 C2 v2 =KC2

    y1 = 2 C1 e 3 tC2 C2 e

    K2 t

    y2 = 4 C1 e 3 tKC2 e

    K2 t

    2. Find the particular solution: 2 C1C2 C2 = 3

    4 C1KC2 = 1

    C1 = 1 2

    , C2 = 1

    y1 = e3 tC2 eK2 t

    y2 = 2 e3 tKeK2 t

    19 of 48

  • Problem: y1' = y1K2 y2 y2' =K2 y1K2 y2 y1(0) = 3 y2(0) = 1

    1. Find the general solution: 1Kλ v1K2 v2 = 0

    K2 v1C K2Kλ v2 = 0

    λ 2 CλK6 = 0

    λ =K3

    v1 =K2 C1 v2 =K4 C1

    λ = 2

    v1 =K2 C2 v2 = C2

    y1 =K2 C1 e K3 tK2 C2 e

    2 t

    y2 =K4 C1 e K3 tCC2 e

    2 t

    2. Find the particular solution: K2 C1K2 C2 = 3

    K4 C1CC2 = 1

    C1 =K 1 2

    , C2 =K1

    y1 = eK3 tC2 e2 t

    y2 = 2 eK3 tKe2 t

    20 of 48

  • Problem: y1' =K5 y1K2 y2 y2' = 2 y1 y1(0) = 1 y2(0) = 1

    1. Find the general solution: K5Kλ v1K2 v2 = 0

    Kλ v2C2 v1 = 0

    λ 2 C5 λC4 = 0

    λ =K1

    v1 =K2 C1 v2 = 4 C1

    λ =K4

    v1 =K2 C2 v2 = C2

    y1 =K2 C1 e KtK2 C2 e

    K4 t

    y2 = 4 C1 e KtCC2 e

    K4 t

    2. Find the particular solution: K2 C1K2 C2 = 1

    4 C1CC2 = 1

    C1 = 1 2

    , C2 =K1

    y1 =KeKtC2 eK4 t

    y2 = 2 eKtKeK4 t

    21 of 48

  • Problem: y1' = 3 y1Ky2 y2' =K2 y1C4 y2 y1(0) = 1 y2(0) = 4

    1. Find the general solution: 3Kλ v1Kv2 = 0

    K2 v1C 4Kλ v2 = 0

    λ 2 K7 λC10 = 0

    λ = 2

    v1 =KC1 v2 =KC1

    λ = 5

    v1 =KC2 v2 = 2 C2

    y1 =KC1 e 2 tKC2 e

    5 t

    y2 =KC1 e 2 tC2 C2 e

    5 t

    2. Find the particular solution: KC1KC2 = 1

    KC1C2 C2 = 4

    C1 =K2, C2 = 1

    y1 = 2 e2 tKe5 t

    y2 = 2 e2 tC2 e5 t

    22 of 48

  • 1st-order, 2 variables: Rea