Pertemuan-10
PD tingkat n Non Homogen
(a0 D n + a1D n −1 + K + an −1a + an ) y = f ( x)
144444
42444444
3
F ( 0) ↵
Non homogen; f ( x) ≠ 0
Untuk Solusi umum PD dari persamaan Non homogen dari fungsi F (0) ⋅ y adalah y1 + y 2
y adalah solusi persamaan Non-homogen → f ( x) ≠ 0
yh adalah solusi persamaan Homogen → f ( x) = 0
yk disebut fungsi khusus / pelengkap / komplementer
Macam-macam bentuk PD. Non-homogen
a.
b.
c.
jika, f ( x) = e kx
f ( x) = cos ax atau sin ax atau kombinasi
f ( x) = a1 x p + a2 x p −1 + K + an (polinom)
kita bahas satu-persatu:
a.
f ( x) = e kx ; k = konstanta
Catatan:
1.
( D − α) e kx = D e kx − α ⋅ e kx
= k e kx − α ⋅ e kx = ( k − α) e kx
( D − α) e kx = ( k − α) e kx
→ D seakan-akan diganti dengan k
2.
( D − α) −1 → invers (D − α)
( D − α ) −1 ⋅ ( D − α ) ⋅ f ( x ) = f ( x )
( D − α ) −1 ⋅ ( D − α ) ⋅ e kx = e kx
( D − α ) −1 ⋅ (k − α ) ⋅ e kx =
e kx
⋅ (k − α )
(k − α )
1
k −α
1
1
=
( D − α) ( k − α)
( D − α ) −1 =
Sehinga kita berkesimpulan D = k atau D dapat diganti k.
Solusi umum PDnya adalah:
F (0) ⋅ y = f ( x)
F (0) ⋅ y = e kx
F −1 (0) = F (0) ⋅ y = F −1 (0) ⋅ e kx
1⋅ y =
yk =
e kx
→ Solusi persamaan Non - homogen
F (k )
e kx
F (k )
yh = Solusi umum PD persamaan homogennya
y = yk + yh
Contoh:
d2y
dy
−5
+ 6 y = 3e − 2 x
2
dx
dx
Persamaan Non-Homogen:
( D 2 − 5 D + 6) y = 3 e -2x ; k = -2 , D=k=-2
( D − 2)( D − 3) y = 3 ⋅ e -2x
y=
3 ⋅ e −2 x
( D − 2)( D − 3)
3 e −2 x
y1 =
=
(−4)(−5)
3
20
e− 2 x
Persamaan homogen
( D 2 − 5D + 6) y = 0
Persamaan Karakteristik :
α 2 − 5α + 6 = 0
(α − 2)(α − 3) = 0
α1 = 2 , α 2 = 3
yh = c1eα x + c2 eα
1
2x
yh = c1e 2 x + c2 e 3 x
Solusi umum PD:
y = yk + yh =
3
20
e −2 x + c1e 2 x + c2e3 x
Catatan:
Jika pada solusi non-homogen pada penyebut = 0 atau k = α n = 0 , dalam hal ini k
tidak boleh langsung mengganti D.
Yang kita lakukan:
e αx ⋅ Dv = ( D − α) e αx
Berlku juga untuk inversnya:
( D − α) −1 ⋅ e kx ⋅ v = e kx ⋅ D −1v
= e kx ⋅ ∫ v dx
Contoh:
( D2 − 5D + 6) y = 3e 2 x ; k = 2
Solusi khusus:
y=
3e 2 x
3e 2 x
1
3e 2 x
=
⋅
= ( D − 2 ) −1 ⋅
−1
( D − 2)( D − 3) ( D − 2) (2 − 3)
Ingat: ( D − α ) −1 e kx ⋅ v = e kx ⋅ D −1 ⋅ v
yk = e 2 x D −1 ⋅ (−3)
yk = e 2 x ∫ − 3 dx = − 3 e 2x ⋅ x
Solusi homogen :
( D 2 − 5D + 6) y = 0
Persamaan Karakteristik :
(α − 2)(α − 3) = 0
α1 = 2 ; α 2 = 3
yh = c1e 2 x + c2e3 x
Solusi umum PD:
y = yk + yh = −3e 2 x x + c1e 2 x + c2e3 x
b.
f ( x) = cos ax atau sin ax atau sin x . cos x
D (cos ax) = -a sin ax
D2 (cos ax) = − a 2 cos ax
D2 seakan-akan sama dengan − a 2
D 2 = −a 2
Contoh:
a.
d2y
+ y = 2 cos 2x , a = 2 , D2 = −a 2 = −4
dx 2
(D
2
)
+ 1 y = 2 cos 2x
2 cos 2x
y=
(D2 + 1)
yk =
2 cos 2x
2
= − cos 2x (solusi khusus)
(−4 + 1)
3
Solusi homogen:
(D
2
+ 1) y = 0
Persamaan Karakteristik :
α 2 +1 = 0
α 2 = −1
α1, 2 = ± − 1
α1,2 = ± i → a = 0 , b = 1
yh = e a. x (A cos bx + B sin bx)
= e0. x (A cos x + B sin x )
= A cos x + B sin x
Solusi umum PD:
y = yk + yh
y=−
b.
2
cos 2x + A cos + B sin x
3
kalau F (0) bukan fungsi kuadrat/pangkatnya ganjil
d 2 y dy
+
+ y = cos x
dx 2 dx
( D2 + D + 1) y = cos x ; a = 1
D 2 = −a 2 = −1
cos x
cos x
cos x D
yk = 2
=
=
⋅
D + D +1 −1+ D +1
D D
D(cos x ) − sin x
yk =
=
= sin x
D2
−1
Solusi homogen :
( D 2 + D + 1) y = 0
Persamaan Karakteristik :
α2 + α + 1 = 0
1
3 ⋅i
−1± 1− 4
=− ±
2
2
2
1⎫
a=− ⎪
2 ⎪ y = e − 12 x ⎛⎜ A cos 3 x + B sin 3 x ⎞⎟
⎬
⎜
3 ⎪ h
2
2 ⎟⎠
⎝
b=
2 ⎪⎭
α1, 2 =
Solusi umum PD:
y = yk + yh
y = sinx + e
c.
− 12 x ⎛
⎞
⎜ A cos 3 x + B sin 3 x ⎟
⎜
2
2 ⎟⎠
⎝
f ( x) = a1 ⋅ x p + a2 x p −1 + K + a p −1 x + a p → (fungsi Polinom)
Contoh:
d2y
dy
+ 3 + 2 y = 3x 2 + 4 x 2 − 2 x + 1
2
dx
dx
2
( D + 3D + 2) y = 3x 3 + 4 x 2 − 2 x + 1
1
yk = 2
⋅ 3x3 + 4 x 2 − 2 x + 1
( D + 3D + 2)
(
2 + 3D + D 2
)
1 3
7
15
− D + D 2 − D3
2 4
8
16
2
D
3
3
9
3
1+ D +
− D − D2 − D2
2
2 − 2
4
4
−
7 2 3 2
D2
3
D
+
D
− D−
4
4
2
2
7 2 21 3 7 4
D + D + D
4
8
8
15 3 7 4
− D − D
8
8
/1
=
Batas pembagian adalah jika hasil pembagian terakhir pang-katnya = pangkat
tertingi dari fungsi polinom persamaan (dalam soal ini pangkat tertinggi = 3)
(
)
(
)
7
15 ⎞
⎛1 3
yk = ⎜ − D + D 2 − D 3 ⎟ 3x 3 + 4 x 2 − 2 x + 1
8
16 ⎠
⎝2 4
(
)
(
)
7
1
3
3x 3 + 4 x 2 − 2 x + 1 − D 3x 3 + 4 x 2 − 2 x + 1 + D 2 3x 3 + 4 x 2 − 2 x + 1
4
8
2
15 3 3
− D 3x + 4 x 2 − 2 x + 1
16
yk =
(
)
(
)
yk =
3 3
1 3
7
15
x + 2 x 2 − x + − 9 x 2 + 8 x − 2 + (18 x + 8) − (18)
2
2 4
8
16
yk =
3 3
1 27 2
3 63
135
x + 2x2 − x + −
x − 6x + + x + 7 −
2
2 4
2 4
8
yk =
3 3 19 2 35
63
x − x + x−
2
4
4
8
1
3
3
7
yk = 1 x 3 − 4 x 2 + 8 x − 7
2
4
4
8
Solusi Homogen :
( D 2 + 3D + 2) y = 0
Persamaan Karakteristik :
α 2 + 3α + 2 = 0
(α + 2)(α + 1) = 0 α1 = −2 α 2 = −1
y2 = c1e −2 x + c2e − x
Solusi umum PD: y = yh + yk
1
3
3
7
y = c1e − 2 x + c2 e − x + 1 x 3 − 4 x 2 + 8 x − 7
2
4
4
8
Contoh-contoh Soal:
1.
d2y
− 4 y = 16 x 2 , tentukan Solusi umum PD!
2
dx
Jawab:
(D
2
)
− 4 y = 16 x 2
1
y= 2
⋅ 16 x 2
D −4
1 D2
− −
4 16
1
− 4 + D2
1−
D2
4
D2
4
D2 D4
−
4 16
⎛ 1 D2 ⎞
⎟⎟ ⋅ 16 x 2 = −4 x 2 − (2)
yk = ⎜⎜ − −
4
16
⎝
⎠
1
1
yk = − 16 x 2 − D 2 16 x 2
4
16
(
(
)
1
(32)
16
yk = −4 x 2 − 2
yk = −4 x 2 −
)
(
)
Solusi Homogen :
(D
2
)
−4 y =0
Persamaan Karakteristik :
α2 − 4 = 0
(α − 2)(α + 2) = 0
α1 = 2 , α 2 = −2
yh = c1e 2 x + c2e −2 x
Solusi umum PD:
y = yh + yk
y = c1e 2 x + c2e −2 x − 4 x 2 − 2
2.
y′′ + y = 2e3 x ; tentukan Solusi umum PD!
Jawab:
dy 2
+ y = 2e3x ; k = 3 , D=k=3
2
dx
( D2 + 1) y = 2e3 x
yk =
2e3 x
2e3 x
2e3 x 1 3 x
=
=
= e (Solusi Khusus)
D 2 + 1 32 + 1 10
5
Solusi Homogen :
( D 2 + 1) y = 0
Persamaan Karakteristik :
α2 + 1 = 0
α1, 2 = ± − 1 = ±i → a = 0 , b = 1
yh = e ax (A cos bx + B sin bx )
yh = A cos x + B sin x
Solusi umum PD:
y = yh + yk
1
y = A cos x + B sin x + e3 x
5
3.
d2y
dy
+2
+ y = 4 sin x ; tentukan Solusi umum PD!
2
dx
dx
Jawab:
( D2 + 2 D + 1) y = 4 sin x ; a = 1 , D 2 = − a 2 = −12 = −1
Solusi khusus:
yk =
4 sin x
D + 2D + 1
2
4 sin x
4 sin x D 2 D(sin x ) 2 cos x
=
⋅ =
=
− 1 + 2D + 1
2D D
−1
D2
= −2 cos x
yk =
Solusi Homogen :
( D 2 + 2 D + 1) y = 0
Persamaan Karakteristik :
α 2 + 2α + 1 = 0
(α + 1)(α + 1) = 0
α1, 2 = −1
yh = c1e − x + c2 ⋅ xe − x
yh = (c1 + c2 x )e − x
Solusi umum PD:
y = yh + yk
y = (c1 + c2 x )e − x − 2 cos x
4.
d3y
d2y
+
4
= e − 4 x ; tentukan Solusi umum PD!
dx 3
dx 2
Jawab:
( D3 + 4 D2 ) y = e −4 x ; k = -4 , D=k=-4
Solusi khusus:
e-4x
e-4x
e −4 x
=
=
;
(D3 + 4 D 2 ) D 2 ( D + 4) 16( D + 4)
-4x
e-4x
e −4 x
−1 e
−1
yk = ( D − 4)
=
⋅ D (1) =
⋅ 1 ⋅ dx
16
16
16 ∫
e- 4x
yk =
⋅ x = 161 xe - 4x
16
yk =
Solusi Homogen :
( D3 + 4D 2 ) y = 0
Persamaan Karakteristik :
α 3 + 4α 2 = 0
α 2 (α + 4 ) = 0
α1,2 = 0 ; α 3 = −4
yh = (c1 + c2 x )eαx + c3eα 3 x
yh = (c1 + c2 x )e 0 x + c3e − 4 x
yh = c1 + c2 x + c3e −4 x
Solusi umum PD:
y = yh + yk
y = c1 + c2 x + c3e −4 x + 161 xe-4x
5.
d 2 y dy
−
− 2 y = 3e 2 x + x 2
2
dx
dx
Jawab:
( D2 − D − 2) y = 3e 2 x + x 2
↑
↑
Type I Type II
I.
( D 2 − D − 2) y = 3e 2 x ; k = 2 , D=k=2
3e 2 x
3e 2 x
3e 2 x
=
=
yk 1 = 2
( D − D − 2) ( D − 2)( D + 1) (2 + 1)( D − 2)
=
3e 2 x
( D − 2) −1 (1) = e 2 x D −1 (1) = e 2 x ∫ (1)dx =e 2 x x
3
yk1 = xe 2 x
II. ( D2 − D − 2) y = x 2
yk 2 =
1
⋅ (x 2 )
D −D−2
2
( )
⎛ 1 D 3 ⎞
yk 2 = ⎜ − + − D 2 ⎟ x 2
⎝ 2 4 8 ⎠
1
1
3
= − x2 + D x2 − D2 x2
2
4
8
( )
( )
( )
1
1
3
= − x 2 + (2 x ) − (2)
2
4
8
1
1
3
yk 2 = − x 2 + x −
2
2
4
−
-2-D+D
2
1 D 3 2
+ − D
2 4 8
1
D D2 2
1+ −
2
2 −
D D2
− +
2
3
D D 2 D3
− −
+
2
4
4 −
3 2 1 3
D − D
4
4
3 2
D
4
Solusi Khusus :
yk = yk1 + yk 2
yk = xe 2 x −
1 2 1
3
x + x−
2
2
4
Solusi Homogen :
( D 2 − D − 2) y = 0
Persamaan Karakteristik :
α2 − α − 2 = 0
α 1 = −1 , α 2 = 2
yh = c1e − x + c2e 2 x
Solusi umum PD:
y = yh + yk
y = c1e − x + c2 e 2 x + xe 2 x −
1 2 1
3
x + x−
2
2
4
PENGERTIAN DARI OPERATOR INVERS
1)
1
adalah operator invers dari f (D) .
f ( D)
2)
1
. f ( x) = ∫ f ( x) dx + C .
D
( dalam hal ini C = 0 ) .
3)
Bila f ( D ) beroperasi pada
1
. f ( x) didapat f ( x) sendiri..
f ( D)
Jadi
4)
y=
1
. f ( x) .
D −α
Boleh ditulis
Atau
(D − α ) y =
f ( x)
dy
− αy = f ( x) , ( adalah P.D. Linear ).
dx
Di mana P ( x) = −α , Q ( x) = f ( x) .
S ( x) = e ∫
y=
y=
1
S ( x)
1
e
−αx
= e∫
−αdx
= e −αx .
[∫ S ( x)Q( x)dx + C ], di mana C = 0
[∫ e
diperoleh :
Maka
P ( x ) dx
−αx
]
f ( x)dx + 0 ,
y = eαx ∫ e −αx f ( x)dx .
1
. f ( x) = eαx ∫ e −αx f ( x)dx .
D −α
5)
y=
1
. f ( x) .
f ( D)
Jika f (D) dapat diuraikan sebagai berikut :
f ( D) = (D − α1 )(D − α 2 )(D − α 3 ) ,
maka
y=
1
. f ( x) .
(D − α1 )(D − α 2 )(D − α 3 )
Untuk menyelesaikan bentuk ini ada 2 cara pemecahan :
I. Dengan cara memisahkan tiap-tiap faktornya.
A3
1
A1
A2
=
+
+
,
(D − α1 )(D − α 2 )(D − α 3 ) D − α1 D − α 2 D − α 3
⎛ A1
A3 ⎞
A2
⎟⎟. f ( x) ,
y = ⎜⎜
+
+
⎝ D − α1 D − α 2 D − α 3 ⎠
y = A1
1
1
1
f ( x) + A2
f ( x) + A3
f ( x) ,
D − α1
D −α2
D − α3
α1 x
Jadi y = A1e
II. Seperti
∫e
−α1 x
f ( x)dx + A2 eα 2 x ∫ e −α 2 x f ( x)dx + A3eα 3 x ∫ e −α 3 x f ( x)dx .
1
. f ( x) = eαx ∫ e −αx f ( x)dx
D −α
diintegrir satu demi satu dari faktor
kesatu sampai faktor ke n, jadi ada n kali integrir.
y=
1
. f ( x) ,
(D − α1 )(D − α 2 )(D − α 3 )
y=
⎡ 1
⎤
f ( x)⎥ ,
.⎢
(D − α 2 )(D − α 3 ) ⎣ (D − α1 )
⎦
y=
{e e
(D − α )(D − α ) ∫
1
1
2
α1 x
3
−α1 x
}
f ( x)dx ,
{
}
{
} ]
y=
⎡ 1
⎤
1
.⎢
eα1 x ∫ e −α1x f ( x)dx ⎥ ,
(D − α 3 ) ⎣ (D − α 2 )
⎦
y=
1
. eα 2 x ∫ e −α 2 x eα1 x ∫ e −α1x f ( x)dx dx ,
(D − α 3 )
[
α3x
Jadi y = e
∫e
−α 3 x
[e
α2x
∫e
−α 2 x
{e ∫ e
α1 x
−α1 x
} ]
f ( x)dx dx dx
Contoh :
Cara I.
Hitunglah solusi khusus dari :
(D
2
− 5 D + 6 )y = e 4 x .
Solusi :
(D
2
− 5 D + 6 )y = e 4 x ,
yk =
yk =
1
.e 4 x ,
D − 5D + 6
2
1
(D − 3)(D − 2)
.e 4 x .
1
(D − 3)(D − 2)
A = (D − 3).
B = (D − 2).
maka
=
A
B
+
,
(D − 3) (D − 2)
1
(D − 3)(D − 2)
=
1
1
=
=1 ,
(D − 2 ) 3 − 2
1
=
1
1
=
= −1 ,
(D − 3) 2 − 3
(D − 3)(D − 2)
1
(D − 3)(D − 2)
=
1
1
1
−1
=
−
+
.
(D − 3) (D − 2) (D − 3) (D − 2)
yk =
⎡ 1
1 ⎤ 4x
.e 4 x = ⎢
−
⎥.e ,
(D − 3)(D − 2)
⎣ (D − 3) (D − 2) ⎦
yk =
1
1
.e 4 x −
.e 4 x ,
(D − 3)
(D − 2 )
1
yk = e3 x ∫ e −3 x .e 4 x dx − e 2 x ∫ e − 2 x .e 4 x dx ,
yk = e3 x ∫ e x dx − e 2 x ∫ e 2 x dx ,
yk = e3 x .e x − e 2 x . 12 ∫ e 2 x d (2 x ) ,
yk = e3 x .e x − 12 e 2 x .e 2 x = e 4 x − 12 e 4 x ,
yk = 12 e 4 x .
Cara II.
Hitunglah solusi khusus dari :
(D
2
− 5 D + 6 )y = e 4 x .
Solusi :
(D
2
− 5 D + 6 )y = e 4 x ,
yk =
yk =
1
.e 4 x ,
D − 5D + 6
2
1
(D − 3)(D − 2)
.e 4 x ,
yk =
⎡ 1
⎤
1
.⎢
e4 x ⎥ ,
(D − 2) ⎣ (D − 3) ⎦
yk =
1
. e3 x ∫ e −3 x .e 4 x dx ,
(D − 2 )
yk =
1
. e3 x e x dx ,
(D − 2 ) ∫
[
[
]
]
[
]
yk =
1
. e3 x .e x ,
(D − 2 )
yk =
1
. e4 x ,
(D − 2 )
[ ]
yk = e 2 x ∫ e − 2 x .e 4 x dx ,
yk = e 2 x ∫ e 2 x dx ,
yk = e 2 x ( 12 )∫ e 2 x d (2 x) ,
yk = 12 e 2 x .e 2 x ,
yk = 12 e 4 x .
6) Operator invers f (D) terhadap e ax .
1
.e ax .
f ( D)
Jika f ( D) = a0 D n + a1 D n −1 + ... + an −1 D + an ,
(
)
maka f ( D)e ax = a0 a n + a1a n −1 + ... + an −1a + an e ax ,
disingkat :
f ( D)e ax = f (a )e ax ,
− − − − − − − − − −− → .
1
f ( D)
1
1
. f ( D)e ax =
. f (a ) e ax ,
f ( D)
f ( D)
e ax =
maka
1
. f (a ) e ax
f ( D)
1
1
. e ax =
. e ax
f ( D)
f (a)
(a bukan akar F(D) = 0 atau f(a) ≠ 0 ) .
7) Jika a akar rangkap satu dari persamaan f ( D) = 0 .
Atau f ( D ) = (D − a ). g (D ) .
Jadi
1
1
=
,
f ( D) (D − a ). g(D )
1
1
e ax =
.e ax ,
(D − a ). g(D )
f ( D)
1
1
e ax
ax
,
e =
⋅
(D − a ) g(a )
f ( D)
⎤
1
1 ⎡ 1
⋅⎢
e ax =
e ax ⎥ ,
f ( D)
g(a ) ⎣ (D − a ) ⎦
[
]
1
1
e ax =
⋅ e ax ∫ e − ax .e ax dx ,
f ( D)
g(a )
[
]
1
1
e ax =
⋅ e ax ∫ dx ,
g(a )
f ( D)
[ ]
1
1
e ax =
⋅ xe ax ,
f ( D)
g(a )
[ ]
1
1
e ax =
⋅ xe ax ,
(D − a ). g(D )
g(a )
Jadi
1
1 x ax
e ax =
⋅ e .
(D − a ). g(D )
g(a ) 1!
8) Jika a akar rangkap dua dari persamaan f ( D) = 0 .
Atau f ( D) = (D − a ) . g (D ) .
2
1
1
e ax =
.e ax ,
2
f ( D)
(D − a ) . g(D )
e ax
1
1
ax
e =
⋅
,
f ( D)
(D − a )2 g(a )
⎤
1
1 ⎡
1
e ax =
e ax ⎥ ,
⋅⎢
2
g (a ) ⎣ (D − a )
f ( D)
⎦
⎤
1
1 ⎡
1
⋅⎢
e ax =
e ax ⎥ ,
f ( D)
g(a ) ⎣ (D − a )(D − a ) ⎦
⎤
1
1
1 ⎡ 1
⋅
e ax =
e ax ⎥ ,
⎢
f ( D)
g(a ) (D − a ) ⎣ (D − a ) ⎦
[
]
1
1
1
e ax =
⋅
e ax ∫ e − ax .e ax dx ,
f ( D)
g(a ) (D − a )
[
]
1
1
1
e ax =
⋅
e ax ∫ dx ,
g(a ) (D − a )
f ( D)
[ ]
1
1
1
e ax =
⋅
xe ax ,
f ( D)
g(a ) (D − a )
[
]
1
1
e ax =
⋅ e ax ∫ e − ax .xe ax dx ,
f ( D)
g(a )
[
]
1
1
e ax =
⋅ e ax ∫ x dx ,
f ( D)
g(a )
[
]
1
1
e ax =
⋅ e ax . 12 x 2 ,
f ( D)
g(a )
1
1 x 2 ax
ax
e =
⋅
e ,
f ( D)
g(a ) 2 !
Jadi
1
(D − a )2 . g(D )
e ax =
1 x 2 ax
⋅
e .
g(a ) 2 !
9) Jika a akar rangkap tiga dari persamaan f ( D) = 0 .
Atau f ( D) = (D − a ) . g (D ) .
3
Jadi
1
1
,
=
f ( D ) (D − a )3 . g(D )
1
1
e ax =
.e ax ,
3
f ( D)
(D − a ) . g(D )
e ax
1
1
e ax =
⋅
,
f ( D)
(D − a )3 g(a )
⎤
1
1 ⎡ 1
e ax =
e ax ⎥ ,
⋅⎢
3
g (a ) ⎣ (D − a )
f ( D)
⎦
⎡ 1
⎤
1
1
1
⋅
e ax =
e ax ⎥ ,
2 ⎢
f ( D)
g(a ) (D − a ) ⎣ (D − a ) ⎦
[
]
1
1
1
e ax =
⋅
e ax ∫ e − ax .e ax dx ,
2
f ( D)
g(a ) (D − a )
[
]
1
1
1
e ax =
⋅
e ax ∫ dx ,
2
f ( D)
g(a ) (D − a )
[ ]
1
1
1
⋅
e ax =
xe ax ,
g(a ) (D − a )2
f ( D)
⎤
1
1
1 ⎡ 1
⋅
e ax =
xe ax ⎥ ,
⎢
f ( D)
g(a ) (D − a ) ⎣ (D − a )
⎦
[
]
1
1
1
e ax =
⋅
e ax ∫ e − ax .xe ax dx ,
f ( D)
g(a ) (D − a )
[
]
1
1
1
e ax =
⋅
e ax ∫ x dx ,
f ( D)
g(a ) (D − a )
[
]
1
1
1
⋅
e ax =
e ax . 12 x 2 ,
f ( D)
g(a ) (D − a )
1
1
1 ⎡ x 2 ax ⎤
⋅
e ax =
⎢ e ⎥,
f ( D)
g (a ) (D − a ) ⎣ 2 !
⎦
1
1 ⎡ ax −ax x 2 ax ⎤
⋅ ⎢e e . e dx ⎥ ,
e ax =
f ( D)
g (a ) ⎣ ∫
2!
⎦
1
1 ⎡ ax x 2 ⎤
ax
⋅ ⎢e
e =
dx ⎥ ,
g (a ) ⎣ ∫ 2 ! ⎦
f ( D)
1
1 ⎡ ax 1 x 3 ⎤
⋅ ⎢e ⋅ ⋅ ⎥ ,
e ax =
g (a ) ⎣
3 2!⎦
f ( D)
1
1 ⎡ ax x 3 ⎤
e ax =
⋅ ⎢e ⋅ ⎥ ,
f ( D)
g (a ) ⎣
3! ⎦
Jadi
1
(D − a )3 . g(D )
e ax =
1 x 3 ax
⋅ e .
g(a ) 3 !
10) Dapat disimpulkan :
1
1 x ax
e ax =
⋅ e ,
(D − a ). g(D )
g(a ) 1!
1
e ax =
1 x 2 ax
⋅
e ,
g(a ) 2 !
1
e ax =
1 x 3 ax
⋅ e ,
g(a ) 3 !
e ax =
1 x n ax
⋅
e
maka
g(a ) n !
(D − a )2 . g(D )
(D − a )3 . g(D )
.
.
.
1
(D − a )n . g(D )
Contoh :
1
x n ax
ax
=
e
e
n!
(D − a )n
Tentukan solusi khusus dari :
(D
3
+ 1)y = 3 + e − x + 5e 2 x .
Solusi :
(D
3
+ 1)y = 3 + e − x + 5e 2 x .
(D
3
+ 1)y = 3 ,
yk1 =
1
.(3) ,
1 + D3
yk1 = (1).(3) = 3 .
(D
3
+1)y = e − x , D = k = -1 ,
yk 2 =
e−x
,
D3 + 1
e−x
yk 2 =
,`
(D + 1) D 2 − D + 1
(
)
yk 2 =
1
1
−x
⋅
(D − D + 1) (D + 1) e ,
yk 2 =
1
1
e− x ,
⋅
(−1) − (−1) + 1 (D + 1)
2
[
]
2
⎤
1 ⎡ 1
yk 2 = ⋅ ⎢
e− x ⎥ ,
3 ⎣ (D + 1) ⎦
[
]
1
yk 2 = ⋅ e − x ∫ e x e − x dx ,
3
[
]
1
yk 2 = ⋅ e − x ∫ dx ,
3
1
1
yk 2 = ⋅ e − x ⋅ x = x e − x .
3
3
(D
3
+ 1)y = 5e 2 x , maka D = k = 2 .
5e 2 x
,
y k3 = 3
D +1
yk3 =
5e 2 x
= 95 e 2 x .
3
(2) + 1
Jadi solusi khusus :
yk = yk1 + yk2 + yk3 ,
yk = 3 + 13 x e − x + 59 e 2 x .
11) Operator invers f ( D) terhadap sinus dan cosinus.
1
sin ax ;
f ( D)
1
cos ax .
f ( D)
Perhatikan :
D sin ax = a cos ax ,
D 2 sin ax = −a 2 sin ax ,
D3 sin ax = −a 3 cos ax ,
D 4 sin ax = a 4 sin ax ,
( )
atau D 2
sin ax = (- a 2 ) sin ax ,
2
2
•
•
•
( )
maka D 2
n
sin ax = (- a 2 ) sin ax
Demikian juga :
D cos ax = −a sin ax ,
D 2 cos ax = −a 2 cos ax ,
n
D 3 cos ax = a 3 sin ax ,
D 4 cos ax = a 4 cos ax ,
( )
atau D 2
cos ax = (- a 2 ) cos ax ,
2
2
•
•
•
( )
maka D 2
Jadi
cos ax = (- a 2 ) cos ax .
n
(D )
sin ax = (- a 2 ) sin ax , dan
(D )
cos ax = (- a 2 ) cos ax .
2 n
2 n
I.
n
n
n
( )
Jika f ( D) = φ D 2 atau polinom dengan eksponen-eksponen genap,maka :
1
1
sin ax =
sin ax , di mana D 2 = − a 2 ,
2
f ( D)
φ (D )
maka
1
1
sin ax =
sin ax , di mana φ (− a 2 ) ≠ 0
2
2
φ (D )
φ (− a )
Analog :
1
1
cos ax =
cos ax .
2
φ (D )
φ (− a 2 )
II.
Jika f ( D) dapat diuraikan seperti :
1
1
, penyebutnya dapat dijadikan dengan
=
2
f ( D) g (D )⋅ h( D)
ekspoenen genap.
h( − D )
h( − D )
1
=
=
,
2
f ( D) g (D )⋅ h( D) ⋅ h(− D) k (D 2 )
maka
Jadi
1
h( − D )
sin ax =
sin ax .
f ( D)
k (D 2 )
1
h( − D )
sin ax =
sin ax , di mana k (−a 2 ) ≠ 0 .
f ( D)
k (− a 2 )
1
h( − D )
cos ax =
cos ax .
f ( D)
k (− a 2 )
Analog
Contoh:
III.
Jika f ( D) = ( D 2 + a 2 ) atau f (−a 2 ) = 0 .
Menurut Rumus Euler : e iax = cos ax + i sin ax , maka
Re(e iax ) = cos ax dan Im(e iax ) = sin ax .
1
1
sin ax = 2
sin ax ,
f ( D)
D + a2
1
1
sin ax = 2
Im(eiax ) ,
f ( D)
D + a2
1
⎡ 1
⎤
sin ax = Im⎢ 2
eiax ⎥ ,
2
f ( D)
⎣D + a
⎦
⎡
⎤
1
1
sin ax = Im⎢
eiax ⎥ ,
f ( D)
⎣ (D + ai )(D − ai ) ⎦
⎡ 1
⎧ 1
⎫⎤
1
sin ax = Im ⎢
eiax ⎬⎥ ,
⎨
f ( D)
⎣ (D + ai ) ⎩ (D − ai ) ⎭⎦
⎡ 1
1
⎧ x iax ⎫⎤
sin ax = Im⎢
⎨ e ⎬⎥ ,
f ( D)
⎣ (ai + ai ) ⎩1! ⎭⎦
{
}
1
⎡ 1
⎤
sin ax = Im⎢
xe iax ⎥ ,
f ( D)
⎣ 2ia
⎦
{
}
1
⎡ 1 i iax ⎤
sin ax = Im⎢
⋅ xe ⎥ ,
f ( D)
⎣ 2ia i
⎦
{
}
1
⎡ i
⎤
sin ax = Im⎢ 2 xe iax ⎥ ,
f ( D)
⎣ 2i a
⎦
⎡ i
⎤
1
xe iax ⎥ ,
sin ax = Im⎢
f ( D)
⎣ 2(− 1)a
⎦
{
}
1
⎡ − ix iax ⎤
e ⎥ ,
sin ax = Im⎢
f ( D)
⎣ 2a
⎦
1
⎡ − ix
(cos ax + i sin ax )⎤⎥ ‘
sin ax = Im⎢
f ( D)
⎣ 2a
⎦
⎡ − ix
⎤
1
i2x
sin ax = Im ⎢
cos ax −
sin ax ⎥ ,
f ( D)
2a
⎣ 2a
⎦
1
x
x
⎡
⎤
sin ax = Im⎢− i
cos ax +
sin ax ⎥ ,
f ( D)
2a
⎣ 2a
⎦
⎡ x
1
⎛ x
⎞⎤
sin ax = Im⎢ sin ax + i⎜ −
cos ax ⎟⎥ ,
f ( D)
⎝ 2a
⎠⎦
⎣ 2a
,
1
x
sin ax = −
cos ax ,
f ( D)
2a
Jadi
1
sin ax = − 2xa cos ax .
D + a2
Analog
1
cos ax =
D + a2
2
2
x
2a
sin ax .
Bukti :
1
1
cos ax = 2
cos ax ,
f ( D)
D + a2
1
1
cos ax = 2
Re(e iax ) ,
2
f ( D)
D +a
1
⎡ 1
⎤
cos ax = Re ⎢ 2
e iax ⎥ ,
2
f ( D)
⎣D + a
⎦
⎡
⎤
1
1
e iax ⎥ ,
cos ax = Re ⎢
f ( D)
⎣ (D + ai )(D − ai ) ⎦
⎡ 1
⎧ 1
⎫⎤
1
cos ax = Re ⎢
e iax ⎬⎥ ,
⎨
f ( D)
⎣ (D + ai ) ⎩ (D − ai ) ⎭⎦
⎡ 1
1
⎧ x iax ⎫⎤
cos ax = Re ⎢
⎨ e ⎬⎥ ,
f ( D)
⎣ (ai + ai ) ⎩1! ⎭⎦
{
}
1
⎡ 1
⎤
cos ax = Re ⎢
xe iax ⎥ ,
f ( D)
⎣ 2ia
⎦
{
}
1
⎡ 1 i iax ⎤
cos ax = Re ⎢
⋅ xe ⎥ ,
f ( D)
⎣ 2ia i
⎦
{
}
1
⎡ i
⎤
cos ax = Re ⎢ 2 xe iax ⎥ ,
f ( D)
⎣ 2i a
⎦
⎡ i
⎤
1
{
xe iax }⎥ ,
cos ax = Re ⎢
f ( D)
⎣ 2(− 1)a
⎦
1
⎡ − ix iax ⎤
e ⎥ ,
cos ax = Re ⎢
f ( D)
⎣ 2a
⎦
1
⎡ − ix
(cos ax + i sin ax )⎤⎥ ‘
cos ax = Re ⎢
f ( D)
⎣ 2a
⎦
⎡ − ix
⎤
1
i2 x
cos ax = Re ⎢
cos ax −
sin ax ⎥ ,
f ( D)
2a
⎣ 2a
⎦
1
x
x
⎡
⎤
cos ax = Re ⎢− i
cos ax +
sin ax ⎥ ,
f ( D)
2a
⎣ 2a
⎦
⎡ x
1
⎛ x
⎞⎤
cos ax = Re ⎢ sin ax + i⎜ −
cos ax ⎟⎥ ,
f ( D)
⎝ 2a
⎠⎦
⎣ 2a
,
1
x
cos ax =
sin ax
f ( D)
2a
1
cos ax =
D + a2
2
13)
x
2a
sin ax
Operator invers f (D) terhadap e ax ⋅ F (x ) .
1
e ax ⋅ F ( x ) ;
f ( D)
F (x ) = fungsi x atau F (x ) = fungsi trigonometri
Perhatikan :
De ax F = De ax ⋅ F + e ax ⋅ DF ,
De ax F = ae ax ⋅ F + e ax ⋅ DF = e ax (D + a )F ,
D 2 e ax F = D{e ax (D + a )F } ,
D 2 e ax F = De ax ⋅ (D + a )F + e ax D.(D + a )F ,
D 2 e ax F = ae ax ⋅ (D + a )F + e ax D.(D + a )F ,
D 2 e ax F = e ax (D + a )(D + a )F = e ax (D + a ) F ,
2
•
•
•
Umumnya : D n e ax F = e ax (D + a ) F .
n
Jika D n = f ( D) , maka
(D + a )n =
Sehingga f ( D)e ax F = e ax f ( D + a ) F .
f ( D + a) .
Seterusnya f ( D)e ax F1 = e ax f ( D + a ) F1 …………………..(*)
Misalkan f ( D + a ) F1 = F atau F1 =
1
F
f ( D + a)
Maka (*) : f ( D)e ax F1 = e ax f ( D + a ) F1 menjadi :
f ( D)e ax ⋅
1
1
⋅F =
e ax ⋅ F
f ( D + a)
f ( D)
e ax ⋅
atau
1
⋅ F = e ax ⋅ F
f ( D + a)
Ada tiga hal penyelesaian solusi
1
e ax cos bx .
f ( D)
Solusi :
Misalkan y =
I.
1
1
e ax cos bx = e ax ⋅
cos bx .
f ( D)
f ( D + a)
(
)
1
1
=
, di mana D 2 = −b 2 ; sehingga g − b 2 ≠ 0 .
f ( D + a) g (D 2 )
Jika
Maka y = e ax ⋅
1
cos bx ,
f ( D + a)
1
e ax
y=e
cos bx =
cos bx .
g D2
g − b2
ax
II.
( )
(
)
1
1
,
=
2
f ( D + a ) g D ⋅ h(D )
Jika
( )
Maka
y=
h(− D )
1
h(− D )
cos bx =
cos bx =
cos bx
2
f ( D + a)
g D ⋅ h(D ) ⋅ h(− D )
g D2 ⋅ k D2
( )
( ) ( )
III.
Jika
1
1
.
= 2
f ( D + a) D + b 2
e ax
e ax
x ax
cos bx = 2
cos bx =
e sin bx
Maka y =
2
f ( D + a)
D +b
2b
14)
Operator invers f ( D) terhadap xF ( x) .
1
xF ( x) , di mana F (x) = fungsi x atau F (x) = fungsi trigonometri.
f ( D)
y = xF .
Perhatikan :
Dy = DxF ,
Dy = Dx ⋅ F + x ⋅ DF ,
Dy = F + x ⋅ DF = xDF + F ,
d
Dy = xDF + ( dD
D )F .
D(Dy ) = D(F + x ⋅ DF ) ,
D 2 y = DF + D( x ⋅ DF ) ,
D 2 y = DF + Dx ⋅ DF + x ⋅ D(DF ) ,
D 2 y = DF + DF + x ⋅ D 2 F ,
D 2 y = 2 DF + x ⋅ D 2 F ,
D 2 y = xD 2 F + 2 DF ,
D 2 y = xD 2 F +
Kesimpulan :
(
d
dD
)
D2 F .
d
Dy = xDF + ( dD
D )F ,
d
D 2 y = xD 2 F + ( dD
D 2 )F ,
•
•
•
Umumnya :
D n y = xD n F +
(
d
dD
)
Dn F .
Jika D n = f ( D) , maka
(D + a )n =
f ( D + a ) dan
Jadi f ( D) y = xf ( D) F + f ' (D )F dan y = xF ,
Sehingga f ( D) xF = xf ( D) F + f ' (D )F
Ambil f ( D) xF1 = xf ( D) F1 + f ' (D )F1 ,
Misalkan f ( D ) F1 = F , maka F1 =
1
F ,
f ( D)
Maka : f ( D) xF1 = xf ( D) F1 + f ' (D )F1 menjadi ,
f ( D) x
1
1
1
F = xf ( D)
F + f ' (D )
F ,
f ( D)
f ( D)
f ( D)
f ( D) x
1
1
F = xF + f ' (D )
F ,
f ( D)
f ( D)
atau xF + f ' (D )
xF = f ( D) x
1
1
F = f ( D) x
F,
f ( D)
f ( D)
1
1
F − f ' (D )
F
f ( D)
f ( D)
− − − − − − − − − − − − − − − − − − − −− →
1
f ( D)
(
d
dD
)
D n F = f ' (D )F .
1
1
1
1
1
xF =
f ( D) x
F−
f ' (D )
F ,
f ( D)
f ( D)
f ( D)
f ( D)
f ( D)
Jadi
1
1
f ' (D )
F
xF = x
F−
f ( D)
f ( D)
{ f ( D)}2