Download - Selesaikan Persamaan Diferensial Berikut: 1. y''- 5y' + 6y = 0 · ⅇλx into the differential equation: ⅆ2 ⅆx2 ⅇλx - 5 ⅆ ⅆx ⅇλx + 6ⅇλx 0 Substitute ⅆ2 ⅆx2 ⅇλx

Transcript

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) 0

ODE names:

Autonomous equation:

y′′(x) -6 y(x) + 5 y′(x)Sturm-Liouville equation:

ⅆxⅇ-5 x y′(x) + 6 ⅇ-5 x y(x) 0

Sturm-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 ⅇ

3 x

Possible intermediate steps:

Solve ⅆ2y(x)

ⅆx2 - 5 ⅆy(x)

ⅆx+ 6 y(x) 0 :

Assume a solution will be proportional to

ⅇλ x for some

constant λ .

Substitute y(x)

λ x into the

ⅇλ x into the

differential equation:

ⅆ2

ⅆx2ⅇλ x - 5

ⅆxⅇλ x + 6 ⅇλ x 0

Substitute ⅆ2

ⅆx2 ⅇλ x

λ2 ⅇλ x

andⅆ

ⅆxⅇλ x

λ ⅇλ x :

λ2 ⅇλ x - 5 λ ⅇλ x + 6 ⅇλ x 0

Factor out ⅇλ x

:

λ2 - 5 λ + 6 ⅇλ x 0

Since ⅇλ x ≠ 0 for

any finite λ

, the zeros must come from the polynomial:λ2 - 5 λ + 6 0

Factor:(λ - 3) (λ - 2) 0

Solve for λ :λ 2 or λ 3

The root λ

2 gives

y1(x)

c1 ⅇ2 x as a

solution, where

c1 is an

arbitrary constant.

2 contoh_materi_wolfram.nb

The root λ

3 gives

y2(x)

c2 ⅇ3 x as a

solution, where

c2 is an

arbitrary constant.The general solution is the sum of the above solutions:

Answer:

y(x) y1(x) + y2(x) c1 ⅇ2 x + c2 ⅇ

3 x

Plots 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

000

y(x)

Initial conditions:

y(0)

y′(0)

More controls

Possible Lagrangian:

ℒ(y′, y, x)1

2ⅇ-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] Cos3 x

2 +

ⅇ-x/2 x C[4] Cos3 x

2 + ⅇ-x/2 C[1] Sin

3 x

2 + ⅇ-x/2 x C[2] Sin

3 x

2

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) 0

Autonomous 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

Solve as a homogeneous linear equation | ▾

Hide steps

y(x) c1 ⅇ-x/2 sin

3 x

2+ c2 ⅇ

-x/2 x sin3 x

2+ c3 ⅇ

-x/2 cos3 x

2+ c4 ⅇ

-x/2 x cos3 x

2

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 to

ⅇλ x for some

constant λ .

Substitute y(x)

ⅇλ x into the

differential equation:

ⅆ4

ⅆx4ⅇλ x + 2

ⅆ3

ⅆx3ⅇλ x + 3

ⅆ2

ⅆx2ⅇλ x + 2

ⅆxⅇλ x + ⅇλ x 0

Substitute ⅆ4

ⅆx4 ⅇλ x

λ4 ⅇλ x

,ⅆ3

ⅆx3 ⅇλ x

λ3 ⅇλ x

,ⅆ2

ⅆx2 ⅇλ x

λ2 ⅇλ x

,

andⅆ

ⅆxⅇλ x

λ ⅇλ x :λ4 ⅇλ x + 2 λ3 ⅇλ x + 3 λ2 ⅇλ x + 2 λ ⅇλ x + ⅇλ x 0

Factor out ⅇλ x

:

λ4 + 2 λ3 + 3 λ2 + 2 λ + 1 ⅇλ x 0

contoh_materi_wolfram.nb 5

Since ⅇλ x ≠ 0 for

any finite λ

, the zeros must come from the polynomial:λ4 + 2 λ3 + 3 λ2 + 2 λ + 1 0

Factor:

λ2 + λ + 12 0

Solve 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 32

both

have muliplicity

2 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 as

solutions, where

c1

,

c2

,

c3

,

and

c4 are

arbitrary 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 identity

ⅇα+ⅈ β ⅇα cos(β) + ⅈ ⅇα sin(β) :

y(x) c1 ⅇ-x/2 cos3 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 cos3 x

2+ ⅈ ⅇ-x/2 sin

3 x

2+ c4 x ⅇ-x/2 cos

3 x

2- ⅈ ⅇ-x/2 sin

3 x

2

Regroup terms:

y(x) (c1 + c2) ⅇ-x/2 cos

3 x

2+ (c3 + c4) ⅇ

-x/2 x cos3 x

2+

ⅈ (c1 - c2) ⅇ-x/2 sin

3 x

2+ ⅈ (c3 - c4) ⅇ

-x/2 x sin3 x

2

Redefine c1 + c2

as c1 ,

ⅈ (c1 - c2) as

c2 ,

c3 + c4 as

c3 , and

ⅈ (c3 - c4) as

c4 , since

these are arbitrary constants:

Answer:

y(x) c1 ⅇ-x/2 cos

3 x

2+ c2 ⅇ

-x/2 sin3 x

2+ c3 ⅇ

-x/2 x cos3 x

2+ c4 ⅇ

-x/2 x sin3 x

2

Plots of sample individual solutions:

x

y

y

y′

y(0) 1y′(0) 0y′′(0) 0

y(3)(0) 0

contoh_materi_wolfram.nb 7

x

y

y

y′

y(0) 0y′(0) 1y′′(0) 0

y(3)(0) 0

x

y

y

y′

y(0) 0y′(0) 0y′′(0) 1

y(3)(0) 0

x

y

y

y′

y(0) 0y′(0) 0y′′(0) 0

y(3)(0) 1

Sample solution family:

10 20 30 40 50x

-1.0

-0.5

0.5

y

(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:

ⅆxⅇ6 x y′(x) - 7 ⅇ6 x y(x) 0

Sturm-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)1

2ⅇ-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

y(0) 0

and

y′(0) 4 :

Assume a solution will be proportional to

ⅇλ x for some

constant λ .

Substitute y(x)

ⅇλ x into the

differential equation:

ⅆ2

ⅆx2ⅇλ x + 6

ⅆxⅇλ x - 7 ⅇλ x 0

Substitute ⅆ2

ⅆx2 ⅇλ x

λ2 ⅇλ x

andⅆ

ⅆxⅇλ x

λ ⅇλ x :

λ2 ⅇλ x + 6 λ ⅇλ x - 7 ⅇλ x 0

Factor out ⅇλ x

:

λ2 + 6 λ - 7 ⅇλ x 0

Since ⅇλ x ≠ 0 for

any finite λ

, the zeros must come from the polynomial:λ2 + 6 λ - 7 0

Factor:(λ - 1) (λ + 7) 0

Solve for λ :λ -7 or λ 1

10 contoh_materi_wolfram.nb

The root λ

-7 gives

y1(x)

c1 ⅇ-7 x as a

solution, where

c1 is an

arbitrary constant.

The root λ

1 gives

y2(x)

c2 ⅇx as a

solution, where

c2 is an

arbitrary constant.The general solution is the sum of the above solutions:

y(x) y1(x) + y2(x) c1 ⅇ-7 x + c2 ⅇ

x

Solve for the unknown constants using the initial conditions:

Compute ⅆy(x)

ⅆx:

ⅆy(x)

ⅆx

ⅆxc1 ⅇ

-7 x + c2 ⅇx

-7 c1 ⅇ-7 x + c2 ⅇ

x

Substitute y(0) 0

into y(x)

ⅇ-7 x c1 + ⅇx c2 :c1 + c2 0

Substitute y′(0) 4

into ⅆy(x)

ⅆx

-7 ⅇ-7 x c1 + ⅇx c2 :-7 c1 + c2 4

Solve the system:

contoh_materi_wolfram.nb 11

Solve the system:

c1 -12

c2 12

Substitute c1 -12

and

c2 12

into

y(x)

ⅇ-7 x c1 + ⅇx c2 :

Answer:

y(x)1

2ⅇ-7 x ⅇ8 x - 1

Plots of the solution:

x

y

y

y′

12 contoh_materi_wolfram.nb