Selesaikan Persamaan Diferensial Berikut: 1. y''- 5y' + 6y = 0 · ⅇλx into the differential...

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Transcript of Selesaikan Persamaan Diferensial Berikut: 1. y''- 5y' + 6y = 0 · ⅇλx into the differential...

  • Selesaikan Persamaan Diferensial Berikut:1. y ''- 5y ' + 6y = 02. y(4)+2y(3)+ 3y '' + 2y ' + y = 03. y '' + 6y ' - 7y = 0; y(0) = 0; y '(0) = 4

    In[1]:= DSolve[y''[x] - 5 y'[x] + 6 y[x] ⩵ 0, y, x]Out[1]= y → Function{x}, ⅇ2 x C[1] + ⅇ3 x C[2]

    In[2]:= DSolve[y''[x]-5y'[x]+6y[x]=0]Input:

    y′′(x) - 5 y′(x) + 6 y(x) 0ODE names:Autonomous equation:y′′(x) -6 y(x) + 5 y′(x)

    Sturm-Liouville equation:ⅆⅆx ⅇ-5 x y′(x) + 6 ⅇ-5 x y(x) 0Sturm-Liouville equation »

    ODE classification:

    second-order linear ordinary differential equation

    Alternate forms:

    y′′(x) 5 y′(x) - 6 y(x)y′′(x) + 6 y(x) 5 y′(x)

    Differential equation solutions: Approximate form

    Solve as a homogeneous linear equation | ▾Hide steps

    y(x) c1 ⅇ2 x + c2 ⅇ3 xPossible intermediate steps:

    Solve ⅆ2y(x)ⅆx2 - 5 ⅆy(x)ⅆx + 6 y(x) 0 :Assume a solution will be proportional toⅇλ x for someconstant λ .Substitute y(x)λ x into the

    http://mathworld.wolfram.com/Sturm-LiouvilleEquation.html

  • ⅇλ x into thedifferential equation:ⅆ2ⅆx2 ⅇλ x - 5 ⅆⅆx ⅇλ x + 6 ⅇλ x 0Substitute ⅆ2ⅆx2 ⅇλ xλ2 ⅇλ xandⅆⅆx ⅇλ xλ ⅇλ x :λ2 ⅇλ x - 5 λ ⅇλ x + 6 ⅇλ x 0

    Factor out ⅇλ x:λ2 - 5 λ + 6 ⅇλ x 0Since ⅇλ x ≠ 0 forany finite λ, the zeros must come from the polynomial:λ2 - 5 λ + 6 0Factor:(λ - 3) (λ - 2) 0Solve for λ :λ 2 or λ 3The root λ2 givesy1(x)c1 ⅇ2 x as asolution, wherec1 is anarbitrary constant.

    2 contoh_materi_wolfram.nb

  • The root λ3 givesy2(x)c2 ⅇ3 x as asolution, wherec2 is anarbitrary constant.The general solution is the sum of the above solutions:

    Answer:

    y(x) y1(x) + y2(x) c1 ⅇ2 x + c2 ⅇ3 xPlots of sample individual solutions:

    x

    y

    y

    y′ y(0) 1y′(0) 0

    x

    y

    y

    y′ y(0) 0y′(0) 1

    Sample solution family:

    0.2 0.4 0.6 0.8 1.0x

    -20-10

    10

    20

    y

    (sampling y(0) and y′(0))

    Interactive differential equation solution plots:

    0.5 1.0 1.5 2.0 2.5 3.0 3.5x

    y

    y(0) 1.y′(0) 1.

    contoh_materi_wolfram.nb 3

  • 0.5 1.0 1.5 2.0 2.5 3.0 3.5

    000

    000

    000y(x)

    Initial conditions:y(0)y′(0)More controls

    Possible Lagrangian:

    ℒ(y′, y, x) 12ⅇ-5 x (y′)2 - 6 ⅇ-5 x y2

    In[3]:= DSolve[y''''[x] + 2 y'''[x] + 3 y''[x] + 2 y'[x] + y[x] ⩵ 0, y, x]Out[3]= y → Function{x}, ⅇ-x/2 C[3] Cos 3 x

    2 +

    ⅇ-x/2 x C[4] Cos 3 x2

    + ⅇ-x/2 C[1] Sin 3 x2

    + ⅇ-x/2 x C[2] Sin 3 x2

    In[4]:= DSolve[y''''[x]+2y'''[x]+3y''[x]+2y'[x]+y[x]⩵0]

    Input:

    y(4)(x) + 2 y(3)(x) + 3 y′′(x) + 2 y′(x) + y(x) 0Autonomous equation:

    3 y(4)(x) -y(x) - 2 y′(x) - 2 y(3)(x) - y(4)(x)Autonomous equation »

    ODE classification:

    higher-order linear ordinary differential equation

    Alternate form:

    y(4)(x) -2 y(3)(x) - 3 y′′(x) - 2 y′(x) - y(x)Differential equation solutions: Approximate form

    Solve homogeneous linear equation |

    4 contoh_materi_wolfram.nb

    http://www.wolframalpha.comhttp://mathworld.wolfram.com/Autonomous.html

  • Solve as a homogeneous linear equation | ▾Hide steps

    y(x) c1 ⅇ-x/2 sin 3 x2

    + c2 ⅇ-x/2 x sin 3 x2

    + c3 ⅇ-x/2 cos 3 x2

    + c4 ⅇ-x/2 x cos 3 x2

    Possible intermediate steps:Solve ⅆ4y(x)ⅆx4 + 2 ⅆ3y(x)ⅆx3 + 3 ⅆ2y(x)ⅆx2 + 2 ⅆy(x)ⅆx + y(x) 0 :Assume a solution will be proportional toⅇλ x for someconstant λ .Substitute y(x)ⅇλ x into thedifferential equation:ⅆ4ⅆx4 ⅇλ x + 2 ⅆ3ⅆx3 ⅇλ x + 3 ⅆ2ⅆx2 ⅇλ x + 2 ⅆⅆx ⅇλ x + ⅇλ x 0Substitute ⅆ4ⅆx4 ⅇλ xλ4 ⅇλ x,ⅆ3ⅆx3 ⅇλ xλ3 ⅇλ x,ⅆ2ⅆx2 ⅇλ xλ2 ⅇλ x,andⅆⅆx ⅇλ xλ ⅇλ x :λ4 ⅇλ x + 2 λ3 ⅇλ x + 3 λ2 ⅇλ x + 2 λ ⅇλ x + ⅇλ x 0Factor out ⅇλ x:λ4 + 2 λ3 + 3 λ2 + 2 λ + 1 ⅇλ x 0

    contoh_materi_wolfram.nb 5

  • Since ⅇλ x ≠ 0 forany finite λ, the zeros must come from the polynomial:λ4 + 2 λ3 + 3 λ2 + 2 λ + 1 0Factor:λ2 + λ + 12 0Solve for λ :λ -1

    2+ ⅈ 3

    2or λ -1

    2+ ⅈ 3

    2or λ -1

    2- ⅈ 3

    2or λ -1

    2- ⅈ 3

    2

    The roots λ- 12±ⅈ 3

    2 both

    have muliplicity2 and give

    y1(x) c1 ⅇ-1/2+ⅈ 3 2 x,

    y2(x) c2 ⅇ-1/2-ⅈ 3 2 x,

    y3(x) c3 ⅇ-1/2+ⅈ 3 2 x x,

    y4(x) c4 ⅇ-1/2-ⅈ 3 2 x x assolutions, wherec1

    ,c2

    ,c3

    ,andc4 arearbitrary constants.The general solution is the of the above solutions

    6 contoh_materi_wolfram.nb

  • The general solution is the sum of the above solutions:

    y(x) y1(x) + y2(x) + y3(x) + y4(x) c1 ⅇ-1/2+ⅈ 3 2 x + c2 ⅇ-1/2-ⅈ 3 2 x + c3 ⅇ-1/2+ⅈ 3 2 x x + c4 ⅇ-1/2-ⅈ 3 2 x x

    Apply Euler's identityⅇα+ⅈ β ⅇα cos(β) + ⅈ ⅇα sin(β) :y(x) c1 ⅇ-x/2 cos 3 x

    2+ ⅈ ⅇ-x/2 sin 3 x

    2+ c2 ⅇ-x/2 cos 3 x

    2- ⅈ ⅇ-x/2 sin 3 x

    2+

    c3 x ⅇ-x/2 cos 3 x2

    + ⅈ ⅇ-x/2 sin 3 x2

    + c4 x ⅇ-x/2 cos 3 x2

    - ⅈ ⅇ-x/2 sin 3 x2

    Regroup terms:

    y(x) (c1 + c2) ⅇ-x/2 cos 3 x2

    + (c3 + c4) ⅇ-x/2 x cos 3 x2

    +ⅈ (c1 - c2) ⅇ-x/2 sin 3 x

    2+ ⅈ (c3 - c4) ⅇ-x/2 x sin 3 x

    2

    Redefine c1 + c2as c1 ,ⅈ (c1 - c2) asc2 ,c3 + c4 asc3 , andⅈ (c3 - c4) asc4 , sincethese are arbitrary constants:

    Answer:

    y(x) c1 ⅇ-x/2 cos 3 x2

    + c2 ⅇ-x/2 sin 3 x2

    + c3 ⅇ-x/2 x cos 3 x2

    + c4 ⅇ-x/2 x sin 3 x2

    Plots of sample individual solutions:

    x

    y

    y

    y′y(0) 1y′(0) 0y′′(0) 0y(3)(0) 0

    contoh_materi_wolfram.nb 7

  • x

    y

    y

    y′y(0) 0y′(0) 1y′′(0) 0y(3)(0) 0

    x

    y

    y

    y′y(0) 0y′(0) 0y′′(0) 1y(3)(0) 0

    x

    y

    y

    y′y(0) 0y′(0) 0y′′(0) 0y(3)(0) 1

    Sample solution family:

    10 20 30 40 50x

    -1.0-0.5

    0.5y

    (sampling y(0), y′(0), y′′(0) and y(3)(0))

    Interactive differential equation solution plots:

    1 2 3 4x

    0

    2

    3

    4

    y

    y(0) 1.y′(0) 1.y′′(0) 1.y(3)(0) 1.

    y(x)

    Initial conditions:y(0)

    8 contoh_materi_wolfram.nb

  • y(0)y′(0)y′′(0)y(3)(0)More controls

    In[5]:= DSolve[{y''[x] + 6 y'[x] - 7 y[x] ⩵ 0, y[0] == 0, y'[0] == 4}, y, x]Out[5]= y → Function{x}, 1

    2ⅇ-7 x -1 + ⅇ8 x

    In[6]:= DSolve[y''[x]+6y'[x]-7y[x]⩵0,y[0]==0,y'[0]==4]Input:{y′′(x) + 6 y′(x) - 7 y(x) 0, y(0) 0, y′(0) 4}ODE names:Autonomous equation:y′′(x) 7 y(x) - 6 y′(x)

    Sturm-Liouville equation:ⅆⅆx ⅇ6 x y′(x) - 7 ⅇ6 x y(x) 0Sturm-Liouville equation »

    ODE classification:

    second-order linear ordinary differential equation

    Alternate forms:{7 y(x) y′′(x) + 6 y′(x), y(0) 0, y′(0) 4}{y′′(x) 7 y(x) - 6 y′(x), y(0) 0, y′(0) 4}Differential equation solutions: Approximate form

    Solve as a homogeneous linear equation | ▾Hide steps

    y(x) 12ⅇ-7 x ⅇ8 x - 1

    Possible intermediate steps:Solve ⅆ2y(x)ⅆx2 + 6 ⅆy(x)ⅆx - 7 y(x) 0, such that

    contoh_materi_wolfram.nb 9

    http://www.wolframalpha.comhttp://mathworld.wolfram.com/Sturm-LiouvilleEquation.html

  • y(0) 0andy′(0) 4 :Assume a solution will be proportional toⅇλ x for someconstant λ .Substitute y(x)ⅇλ x into thedifferential equation:ⅆ2ⅆx2 ⅇλ x + 6 ⅆⅆx ⅇλ x - 7 ⅇλ x 0Substitute ⅆ2ⅆx2 ⅇλ xλ2 ⅇλ xandⅆⅆx ⅇλ xλ ⅇλ x :λ2 ⅇλ x + 6 λ ⅇλ x - 7 ⅇλ x 0

    Factor out ⅇλ x:λ2 + 6 λ - 7 ⅇλ x 0Since ⅇλ x ≠ 0 forany finite λ, the zeros must come from the polynomial:λ2 + 6 λ - 7 0Factor:(λ - 1) (λ + 7) 0Solve for λ :λ -7 or λ 1

    10 contoh_materi_wolfram.nb

  • The root λ-7 givesy1(x)c1 ⅇ-7 x as asolution, wherec1 is anarbitrary constant.

    The root λ1 givesy2(x)c2 ⅇx as asolution, wherec2 is anarbitrary constant.The general solution is the sum of the above solutions:y(x) y1(x) + y2(x) c1 ⅇ-7 x + c2 ⅇxSolve for the unknown constants using the initial conditions:

    Compute ⅆy(x)ⅆx :ⅆy(x)ⅆx ⅆⅆx c1 ⅇ-7 x + c2 ⅇx -7 c1 ⅇ-7 x + c2 ⅇxSubstitute y(0) 0into y(x)ⅇ-7 x c1 + ⅇx c2 :c1 + c2 0Substitute y′(0) 4into ⅆy(x)ⅆx-7 ⅇ-7 x c1 + ⅇx c2 :-7 c1 + c2 4Solve the system:

    contoh_materi_wolfram.nb 11

  • Solve the system:c1 - 12c2 12Substitute c1 - 12andc2 12 intoy(x)ⅇ-7 x c1 + ⅇx c2 :Answer:

    y(x) 12ⅇ-7 x ⅇ8 x - 1

    Plots of the solution:

    x

    y

    y

    y′

    12 contoh_materi_wolfram.nb

    http://www.wolframalpha.com