Second Order Nonhomogeneous Linear Differential xchen/teach/ode/2nd_ord_nonhomog...Second Order Nonhomogeneous Linear ... are solutions of the homogeneous equation: a2y ... Second

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  • Second Order Nonhomogeneous Linear

    Differential Equations with Constant

    Coefficients:

    the method of undetermined coefficients

    Xu-Yan Chen

  • Second Order Nonhomogeneous Linear DifferentialEquations with Constant Coefficients:

    a2y(t) + a1y

    (t) + a0y(t) = f(t),

    where a2 6= 0, a1, a0 are constants, and f(t) is a given function(called the nonhomogeneous term).

    General solution structure:

    y(t) = yp(t) + yc(t)

    where yp(t) is a particular solution of the nonhomog equation,and yc(t) are solutions of the homogeneous equation:

    a2y

    c (t) + a1y

    c(t) + a0yc(t) = 0.

    The characteristic roots: a22 + a1 + a0 = 0

    The complementary solutions yc(t).

  • Second Order Nonhomogeneous Linear DifferentialEquations with Constant Coefficients:

    a2y(t) + a1y

    (t) + a0y(t) = f(t),

    where a2 6= 0, a1, a0 are constants, and f(t) is a given function(called the nonhomogeneous term).

    General solution structure:

    y(t) = yp(t) + yc(t)

    where yp(t) is a particular solution of the nonhomog equation,and yc(t) are solutions of the homogeneous equation:

    a2y

    c (t) + a1y

    c(t) + a0yc(t) = 0.

    The characteristic roots: a22 + a1 + a0 = 0

    The complementary solutions yc(t).

  • Second Order Nonhomogeneous Linear DifferentialEquations with Constant Coefficients:

    a2y(t) + a1y

    (t) + a0y(t) = f(t),

    where a2 6= 0, a1, a0 are constants, and f(t) is a given function(called the nonhomogeneous term).

    General solution structure:

    y(t) = yp(t) + yc(t)

    where yp(t) is a particular solution of the nonhomog equation,and yc(t) are solutions of the homogeneous equation:

    a2y

    c (t) + a1y

    c(t) + a0yc(t) = 0.

    The characteristic roots: a22 + a1 + a0 = 0

    The complementary solutions yc(t).

    What is this note about? The Method ofUndetermined Coefficients:

  • Second Order Nonhomogeneous Linear DifferentialEquations with Constant Coefficients:

    a2y(t) + a1y

    (t) + a0y(t) = f(t),

    where a2 6= 0, a1, a0 are constants, and f(t) is a given function(called the nonhomogeneous term).

    General solution structure:

    y(t) = yp(t) + yc(t)

    where yp(t) is a particular solution of the nonhomog equation,and yc(t) are solutions of the homogeneous equation:

    a2y

    c (t) + a1y

    c(t) + a0yc(t) = 0.

    The characteristic roots: a22 + a1 + a0 = 0

    The complementary solutions yc(t).

    What is this note about? The Method ofUndetermined Coefficients: a method of finding yp(t), whenthe nonhomog term f(t) belongs a simple class.

  • Second Order Nonhomogeneous Linear DifferentialEquations with Constant Coefficients:

    a2y(t) + a1y

    (t) + a0y(t) = f(t),

    where a2 6= 0, a1, a0 are constants, and f(t) is a given function(called the nonhomogeneous term).

    General solution structure:

    y(t) = yp(t) + yc(t)

    where yp(t) is a particular solution of the nonhomog equation,and yc(t) are solutions of the homogeneous equation:

    a2y

    c (t) + a1y

    c(t) + a0yc(t) = 0.

    The characteristic roots: a22 + a1 + a0 = 0

    The complementary solutions yc(t).

    What is this note about? The Method ofUndetermined Coefficients: a method of finding yp(t), whenthe nonhomog term f(t) belongs a simple class.

    Main Idea: Set up a trial function yp(t), by copying thefunction form of f(t).

  • Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

  • Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

    General solutions y(t) = yc(t) + yp(t).

  • Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

    General solutions y(t) = yc(t) + yp(t).

    Find complementary solutions yc(t):

  • Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

    General solutions y(t) = yc(t) + yp(t).

    Find complementary solutions yc(t):

    3yc + y

    c 2yc = 0 (the corresponding homog eq)

    32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

    2

    3t

  • Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

    General solutions y(t) = yc(t) + yp(t).

    Find complementary solutions yc(t):

    3yc + y

    c 2yc = 0 (the corresponding homog eq)

    32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

    2

    3t

    To find yp(t), set the trial function

    yp(t) = ae4t (form copied from f(t) = 10e4t)

    where a is the undetermined coefficient.

  • Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

    General solutions y(t) = yc(t) + yp(t).

    Find complementary solutions yc(t):

    3yc + y

    c 2yc = 0 (the corresponding homog eq)

    32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

    2

    3t

    To find yp(t), set the trial function

    yp(t) = ae4t (form copied from f(t) = 10e4t)

    where a is the undetermined coefficient.

    Substitute yp(t) in the nonhomog eq:

    3(ae4t) + (ae4t) 2ae4t = 10e4t

    = 3(16ae4t) + (4ae4t) 2ae4t

    = 50ae4t

  • Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

    General solutions y(t) = yc(t) + yp(t).

    Find complementary solutions yc(t):

    3yc + y

    c 2yc = 0 (the corresponding homog eq)

    32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

    2

    3t

    To find yp(t), set the trial function

    yp(t) = ae4t (form copied from f(t) = 10e4t)

    where a is the undetermined coefficient.

    Substitute yp(t) in the nonhomog eq:

    3(ae4t) + (ae4t) 2ae4t = 10e4t

    = 3(16ae4t) + (4ae4t) 2ae4t

    = 50ae4t

    Compare the coefficients of the two sides:

    50a = 10 a =1

    5

  • Example 1: Solve 3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

    General solutions y(t) = yc(t) + yp(t).

    Find complementary solutions yc(t):

    3yc + y

    c 2yc = 0 (the corresponding homog eq)

    32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

    2

    3t

    To find yp(t), set the trial function

    yp(t) = ae4t (form copied from f(t) = 10e4t)

    where a is the undetermined coefficient.

    Substitute yp(t) in the nonhomog eq:

    3(ae4t) + (ae4t) 2ae4t = 10e4t

    = 3(16ae4t) + (4ae4t) 2ae4t

    = 50ae4t

    Compare the coefficients of the two sides:

    50a = 10 a =1

    5 yp(t) =

    1

    5e4t

  • Example 1 (continued): Solve3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

    Combine yc and yp to get

    Gen Sols of Nonhomg Eq: y(t) =1

    5e4t + C1e

    t + C2e2

    3t.

  • Example 1 (continued): Solve3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

    Combine yc and yp to get

    Gen Sols of Nonhomg Eq: y(t) =1

    5e4t + C1e

    t + C2e2

    3t.

    Use initial conditions:

    y(0) = 1 15

    + C1 + C2 = 1

    y(t) = 45e4t C1e

    t + 23C2e

    2

    3t, y(0) = 3 4

    5 C1 +

    2

    3C2 = 3

  • Example 1 (continued): Solve3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

    Combine yc and yp to get

    Gen Sols of Nonhomg Eq: y(t) =1

    5e4t + C1e

    t + C2e2

    3t.

    Use initial conditions:

    y(0) = 1 15

    + C1 + C2 = 1

    y(t) = 45e4t C1e

    t + 23C2e

    2

    3t, y(0) = 3 4

    5 C1 +

    2

    3C2 = 3

    Solve this:

    {

    C1 = 9

    5

    C2 =3

    5

  • Example 1 (continued): Solve3y + y 2y = 10e4t, y(0) = 1, y(0) = 3.

    Combine yc and yp to get

    Gen Sols of Nonhomg Eq: y(t) =1

    5e4t + C1e

    t + C2e2

    3t.

    Use initial conditions:

    y(0) = 1 15

    + C1 + C2 = 1

    y(t) = 45e4t C1e

    t + 23C2e

    2

    3t, y(0) = 3 4

    5 C1 +

    2

    3C2 = 3

    Solve this:

    {

    C1 = 9

    5

    C2 =3

    5

    The solution of the initial value problem:

    y(t) =1

    5e4t

    9

    5et +

    3

    5e

    2

    3t.

  • Nonhomogeneous Linear Equations:

    a2y(t) + a1y

    (t) + a0y(t) = f(t),

    Towards the Rules of Setting Up the Trial Function:

    f(t) yp(t)

    kert Aert

    (to be continued)

  • Example 2: Solve 3y + y 2y = 8te2t.

  • Example 2: Solve 3y + y 2y = 8te2t.

    Complementary solutions yc(t):

    32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

    2

    3t

  • Example 2: Solve 3y + y 2y = 8te2t.

    Complementary solutions yc(t):

    32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

    2

    3t

    To find yp(t), set the trial function

    yp(t) = Ate2t.

  • Example 2: Solve 3y + y 2y = 8te2t.

    Complementary solutions yc(t):

    32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

    2

    3t

    To find yp(t), set the trial function

    yp(t) = Ate2t.

    Substitute yp(t) in the nonhomog eq:

    8te2t = 3(Ate2t) + (Ate2t) 2Ate2t

    = 3(4Ae2t + 4Ate2t) + (Ae2t 2Ate2t) 2Ate2t

    = 11Ae2t + 8Ate2t

  • Example 2: Solve 3y + y 2y = 8te2t.

    Complementary solutions yc(t):

    32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

    2

    3t

    To find yp(t), set the trial function

    yp(t) = Ate2t.

    Substitute yp(t) in the nonhomog eq:

    8te2t = 3(Ate2t) + (Ate2t) 2Ate2t

    = 3(4Ae2t + 4Ate2t) + (Ae2t 2Ate2t) 2Ate2t

    = 11Ae2t + 8Ate2t

    Compare the coefficients of the two sides:{

    11A = 08A = 8

    Impossible!

  • Example 2: Solve 3y + y 2y = 8te2t.

    Complementary solutions yc(t):

    32 + 2 = 0 1 = 1, 2 = 2/3 yc = C1et + C2e

    2

    3t

    To find yp(t), set the trial function

    yp(t) = Ate2t.

    Substitute yp(t) in the nonhomog eq:

    8te2t = 3(Ate2t) + (Ate2t) 2Ate2t

    = 3(4Ae2t + 4Ate2t) + (Ae2t 2Ate2t) 2Ate2t

    = 11Ae2t + 8Ate2t

    Compare the coefficients of the

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