tugas matkim secara keseluruhan
TRANSCRIPT
Natural science ( ILMU ALAM )
Mempelajari zat dan energi :
1. Fisika : phenomenal
2. Astronomi
3. Kimia : detail reaktif
4. Geologi
5. Biologi : implemantasi hidup
Esensinya
Mekanika : sebagai pemberi makna fisis (bergerak)
Logika (alat komunikasi), terdiri dari : induktif ( statistik ) dan deduktif ( matematika )
Hukum laju reaksi adalah
A + B P
Hukum laju reaksi :
V = k [A]x [B}y
V = Ae-EaKbt [A]x [B}y
V = c σ* e-EaKbt [A]x [B}y
Untuk mencari pengaruh dari salah satu faktor, maka faktor yang lain dibuat tetap.
Persamaan Diferensial
1.dydx
+ sin x = 0
2.d2 ydx 2
+ y x = 0
3.d2wdx 2
+ h
mEy = 0
Untuk menyelesaikan Persamaan Dasar, yaitu dengan cara :
1. Variabel terpisah ( memisahkan variabel lalu mengintegralkannya )
F1 (x) . G2(y) dx + G1(y) . F2(x) dy = 0
Maka kedua ruas dikalikan dengan 1
g2( y) f 2(x)
f 1(x )f 2(x )
dx+g1( y)g2( y)
dy=0
∫ f 1(x)f 2(x)
dx+∫ g1( y )g2( y )
dy=∫0
F(x) + G(x) = c
Contoh :
1. x3 dx+ ( y+1 )2dy=0 variabel sudah terpisah maka langsung diintegralkan
∫ x3 dx+∫ ( y+1 )2dy=∫ 0
14
x4+ 13
( y+1 )2 dy=C agar bulat maka dikali 12
3 x4+4 ( y+1 )3=12C
3 x4+4 ( y+1 )3=C
2.dx
sin5 x+ dy
(1−sinx )=0
Dikali dengan (sin5 x ) (1−sinx )
(1−sinx )dx+sin5 x dy=0
∫ (1−sinx ) dx+∫ sin5 x dy=∫0
∫ dx+∫ dcosx+∫ (sin2 y )2sin y dy=∫ 0
x+cos x+∫ (1−cos2 y )2 dcos y=C
x+cos x+∫ (1−2cos2 y+cos4 y ) d cos y=C
x+cos x+cos y−23
cos3 y+ 15
cos5 y=C
15 x+15 cos x+15 cos y−10 cos3 y+3 cos y=C
Contoh soal
1. (1+x ) ydx+(1− y ) x dy=0 dikali dengan 1xy
(1+x ) (1+x )x
dx+(1− y ) (1− y )y
dy=0 diintegralkan
∫ (1+x ) 1x
dx+∫ (1− y ) 1y
dy=∫ 0
Konstanta yang baru
Konstanta yang baru
∫ 1x
dx+dx+n∫ 1y
dy−dy=∫0
ln x+x+ ln y− y=C
Pembuktian :
Mendifferensialkan hasilnya
d ln x+dx+d ln y−dy=dC
1x
dx+dx+ 1y
dy−dy=0
1+ x
xdx+ 1− y
ydy=0 dikali xy
(1+x ) ydx+(1− y ) x dy = 0 (TERBUKTI)
RUMUS INTEGRAL
1. ∫ eu du=eu+C
2. ∫ au du= au
ln au+C
3. ∫undu=¿ un+1
n+1+C ¿
4. ∫ lnu du=u ln u−u+C
5. ∫ duu ln u
=ln ( lnu )+C
6. ∫ueu du=eu(¿u−1)+C ¿
7. ∫unlnudu=u
n+1[ ¿un+1
−1
(n+1)2 ]+C
A x P
Ketika t = 0 maka A = a
t = t maka A = ( a-x )
maka persamaan laju reaksinya adalah :
−dAdt
=dpdt
=k [ A ]1
−d (a−x )dt
=k ( a−x )
−d (a−x)(a−x )
=k dt
∫0
x −d (a−x )( a−x )
=∫0
t
k dt
−ln ( a−x )∫0
x
¿kt∫0
t
❑
−ln ( a−x )+ln a=kt
ln a−ln ( a−x )=kt
lna
a−x=kt , maka grafiknya :
Berkurang sebesar x tiap waktu ( t )
lna
a−x
------------------
t
jika t=t 12 atau waktu paroh
sehingga a ~ a/2 , sehingga persamaan menjadi
lnaa2
=k t 12
ln 2=¿k t 12
¿
0,693=k t 12
t 12=0,693
k , satuan waktu sedngkan didapat dari grafik.
lna
(a−x )=kt
a(a−x )
=ekt
(a−x )=a e−kt
At = ae−kt
A = Ao e− tℷ
ℷ=konstanta peluruhan
At = peluruhan Radioaktif
SOAL
1. A + B x P
t = 0 A=a, B=a
t=t A=a-x , B=a-x
maka persamaan laju reaksinya
−dAdt
=−dBdt
=dPdt
=k [ A ] [B]
−d (a−x)dt
=k ( a−x )(a−x )
∫−d (a−x )(a−x )2 =∫ k dt
−∫(a−x )−2 d ( a−x )=∫ k dt
−1−2+1
∫0
x
¿k t∫0
t
❑
1(a−x )∫0
x
¿kt∫0
t
❑
1(a−x )
−1a=kt
a−a+xa (a−x )
=kt
xa(a−x )
=kt
2. A + B + C x P
t=0 A=B=C=a
t=t A=B=C=a-x
maka persamaan laju reaksinya :
−dAdt
=−dBdt
=−dCdt
=k [ A ] [B ] [C ]
−d (a−x)dt
=k ( a−x ) (a−x )(a−x )
−d (a−x )dt
=k ( a−x )3
−d (a−x)(a−x)3 =k dt
∫−d (a−x )(a−x )3 =∫ k dt
−(−12
(a−x )−2)∫0
x
¿k t∫0
t
❑
12(a−x)2∫
0
x
¿kt∫0
t
❑
1
2(a−x)2− 1
2a2=kt
PERSAMAAN DIFERENSIAL HOMOGEN TINGKAT 1
f (x,y) = f (ƛx1 . ƛy) = ƛ f (x1.y)
Soal
(x2 +y2) dx = 2xy dy
x ƛ
[ ƛ x2 + ƛ y2] dx = 2xy dy
[(ƛx)2 + (ƛy)2] dx = 2 (ƛx) (ƛy) dy
[ƛ2 x2 + ƛ2y2] dx = 2 (ƛ2 xy) dy
ƛ2 (x2 + y2) dx = ƛ2 2xy dy
(x2 + y2) dx = 2xy dy
: x2
(1 + y2/ x2) dx = 2y/x dy
Misal : y = u x dy = u dx + x du
u = y/x
(1 + u2) dx = 2u (u dx + x du)
(1 + u2) dx = 2u2 dx + 2 ux du
(1 + u2 ) dx - 2u2 dx - 2 ux du = 0
(1 + u2 - 2u2) dx – 2 ux du = 0
[(1 - u2 ) dx - 2 ux du] = 0 ditukar
dx / x – 2u / 1-u2 du = 0
ʃ dx / x - ʃ 2u / 1-u2 du = 0
ln x + ln (1 - u2 ) = ln C
ln x (1 - u2 ) = ln C
x (1 - u2 ) = C
x (1- (y/x)2) = C
{x( 1- y2/x2) = C} : x2
x / x2 (x2 - y2) = C
[x2 - y2 / x] = C dikali x
x2 - y2 = C x
Contoh Soal
1. (x3 + y3) dx – 3xy2 dy = 0
[(x3 + y3) dx = 3xy2 dy] : x3
(1 + y3 / x3 ) dx = 3 y2 / x2 dy
Misal :
y = u x ; dy = u dx + x du
u =y /x
(1 + u3) dx = 3 u2 (u dx + x du)
(1 + u3) dx = 3 u3 dx + 2 u2 x du
(1 + u3) dx - 3 u3 dx - 3 u2 x du = 0
(1 + u3 - 3 u3) dx - 3 u2 x du = 0
[(1-2u3) dx – 3u2x du = 0] ditukar
dx/x – 3u2 / (1-2u3) du = 0
ʃ dx / x -ʃ 3u2 / (1-2u3) du = ʃ 0
ln x + ½ ln (1-2u3) = ln C
2 ln x + ln (1-2u3) = ln C
ln x2 (1-2u3) = ln C
x2 (1-2u3) = C
x2 (1-2 (y/ x)3 ) =C
{ x2 (1-2 y3/ x3 ) = C} dibagi x3
x2 / x3 (x3 – 2y3) = cx3
[ 1/x (x3 – 2y3) = C] dikali x
x3 – 2y3 = Cx
2. (2x + 3y) dx + (y – x) dy = 0
[ (2x + 3y) dx = - (y – x) dy ] dibagi x
(2 + 3y / x) dx = -y /x + 1 dy
Misal :
y = ux ; dy = u dx + x du
u = y/x
(2 + 3u) dx = -u +1 (u dx + x du )
(2 + 3u) dx = -u +1 (u dx + x du )
(2 + 3u) dx = -u2 dx – u x du + u dx + x du
(2 + 3u) dx + u2 dx + u x du - u dx - x du = 0
(2 + 3u + u2 – u) dx + (ux – x) du = 0
dx / ux – x + du / 2+3u+u2 = 0
ʃ dx / ux-x + ʃ du / 2+3u+u2 = ʃ 0
Persamaan Differensial Linear Tingkat 1
dydx
+P (x) y=g(x )
ye∫p ( x ) dx → y = U.V
dy = U.dV + V.dUdydx
e∫ p ( x ) dx = y.de∫p ( x ) dx + e∫ p ( x ) dx dy
dx
= y p(x)e∫ p ( x ) dx + e∫ p ( x ) dx dydx
= e∫ p ( x ) dx [ y p ( x )+ dydx ]
= e∫ p ( x ) dx [ dydx
+ p(x ) y ]= e∫ p ( x ) dx
Q(x )
= Q ( x ) . e∫ p ( x ) dxdx
∫ dye∫p ( x )dx = ∫Q( x)e∫ p ( x ) dx
dx
y e∫p ( x ) dx = ∫Q( x)e∫ p ( x ) dx
dx+c
Contoh soal :dydx
− y=2ex
y. e∫ p ( x ) dx = ∫Q ( x ) e∫ p (x ) dxdx+c
ye−∫ dx=∫2ex . e
−∫dxdx+c
ye∫−x=∫2ex . e− xdx+c
ye− x = ∫ dx+c
ye− x=2x+c
y=2x ex+c ex
dydx
=2 [x ex+ex ]+ c ex
dydx
− y=2ex
2 ( x ex+ex)+C ex−(2 x ex+C ex )=2e x
2 xex+2ex−2 x ex−C ex=2e x
2ex = 2ex
Soal
1.dydx
+ 3x2y = x . e− x3
Jawab :
1.dydx
+ 3x2y = x . e− x3
y . e∫p ( x ) dx = ∫Q( x) . e∫ p ( x ) dxdx + C
y . e∫3 x2dx = ∫ x . e−x 3 . e∫
3 x2dxdx+C
y . ex 3 = ∫ x . e−x 3 . ex 3 dx+C
y . ex 3 = 12
x2 + C
y = 12
x2 . e− x3+C .e− x3
PERSAMAAN DIFFERENSIAL BERNOULLI
dydx
+ p ( x ) y=Q(x ) ynmisal :
y−n dydx
+ p ( x ) y(−n+ 1)=Q(x ) u= y(−n+1)
1(−n+1)
dydx
+p (x ) u=p (x) dudx
= (-n+1)y−n dy
dx
y−n dydx
= 1(−n+1)
dydx
contoh
dydx
− y=x y5
Jawab :
dydx
− y=x y5misal
y−n dydx
− y−4=x y−4=u
-4dudx
−u=xdudx
=−4 y−5 dydx
dudx
+4u=−4 x -4dudx
= y−5 dydx
u . e∫4dx = ∫−4 x . e∫
4dx. dx+c
u . e4x=∫−4.e4x dx+c
u . e4x=−∫ x d e4x+c
¿ [x e4x−∫ e4x dx ]+c
u . e4x=−[ xe4 x−14
e4 x]+c
:e4 x
u=−x+ 14+c . e−4 x
y−4=−x+ 14+c . e−4 x
Pembuktian:
dy−4=d(−x+ 14+c .e−4 x)
−4 y−5 dy=(−1−4 c . e−4 x ) dx
−4 y−5 dydx
=(−1−4 c . e−4 x )
: -4
y−5 dydx
=( 14+c . e−4 x)
( 14+c . e−4 x)— x+( 1
4+c . e−4x )=x
x=x (Terbukti)
Soal
1.dydx
+xy=x3 y3
2. Sin y dydx
=cos y (1−xcos y )
3.dydx
+ y tg x= y3 sec6 x
4.dydx
+ yctg x= y2 cos2 x
Jawaban
1.dydx
+xy=x3 y3
y−3 dy
dx + xy−2 =x3
−du2dx
+ xu = x3
dudx
−2 xu=−2x3
u x ∫−2 xdx= ∫−2 x3e ∫−2x3 dx
ue− x2= ∫ −2 x3e− x2 dx + c
= ∫ −2 x2 x e−x 2 dx + c
=∫x2 de− x2 + c
=x2 e−x 2− ∫ e− x2 dx2
=x2 e−x 2− ∫ 2 x . e−x 2 dx
= x2 e−x 2+ ∫ de−x 2
= x2 e−x 2+e−x 2 + c
U = x2+ 1 + cex 2
y−3 = x2 + 1 + cex 2
Misal :
y−2 = u
dudx
= -2 y−3 dy
dx
PERSAMAAN DIFFERENSIAL EKSAK
F(x,y) = C
dF(x,y)= (∂F∂ x
)y dx + (∂F∂ y
)x dy
0 = (∂F∂ x
)y dx + (∂F∂ y
)x dy
(∂F∂ x
)y = M(x,y) (∂F∂ y
)x = N(x,y)
(∂F∂ x
)y dx + (∂F∂ y
)x dy = 0
M(x,y) dx + N(x,y) dy = 0
(∂ M∂ y
)x = (∂N∂ x
)y
(∂F∂ x
)y = M(x,y)
∫ dF = ∫M (x,y) dx + h(y)
F(x,y) = ∫M (x,y) dx + h(y)
(∂F∂ x
) = ∂
∂ y ∫M (x,y) dx + d
h( y )dy
N(x,y) = ∂
∂ y ∫M (x,y) dx +
dh( y )dy
dh(y) = [N(x,y) - ∂
∂ y ∫M (x,y)dx]dy
h(y) = ∫¿¿N(x,y) - ∂dy∫M (x,y)dx]dy
F(x,y) = ∫M (x,y)dx + ∫N (x,y) - ∂
∂ y ∫M (x,y)dx]dy
nRV
dT - nRTV 2
dV = 0
M(T,V) = nRV
N(T,V) = - nRTV 2
(∂ M∂V
)T = - nRV 2
(∂N∂T
)V = -nRV 2
(∂ M∂V
)T = (∂N∂T
)V
(∂F∂T
)V = M(T,V)
∫ dF = ∫ nRV
dT + h(V)
F(T,V) = nRTV
+ h(V)
N(T,V)
(∂ F∂V
)T = - nRTV 2
+ dh (V )
dV
- nRTV 2
= - nRTV 2
+ dh (V )
dV
0 = dh (V )
dV = h(V) = 0
F(T,V) = nRTV
+ 0
P = nRTV
→ PV = nRT
SOAL
1. (3e3xy-2x)dx + e3xdy = 0
2. (y2exy2 + 4x3)dx + (2xyexy2 – 3y2)dy = 0
JAWABAN
1. (3e3xy-2x)dx + e3xdy = 0
M(x,y) = 3e3xy-2x N(x,y) = e3x
(∂ M∂ y
)x = 3e3x (∂N∂ x
)y = 3e3x
(∂ M∂ y
)x = (∂N∂ x
)y
(∂F∂ x
)y = M(x,y) (∂F∂ y
)x = M(x,y)
∫ dF = ∫M (x,y)dx + h(y)
∫ dF = ∫¿¿3e3xy-2x)dx + h(y)
F(x,y) = 3/3e3xy-x2 + h(y)
F(x,y) = e3xy-x2 + h(y)
Mencari nilai h(y) dengan cara mendiferensial secara parsial persamaan tersebut
terhadap x :
(∂F∂ y
)x = e3x + dh( y )
dy
e3x = e3x + dh( y )
dy
0 = dh( y )
dy
h(y) = 0
Jadi, hasilnya adalah....
F(x,y) = e3xy-x2 + h(y)
F(x,y) = e3xy-x2
2. (y2exy2 + 4x3)dx + (2xyexy2 – 3y2)dy = 0
M(x,y) = y2exy2 + 4x3 N(x,y) = 2xyexy2 – 3y2
= UdV + VdU = UdV + VdU
= y2.2xyexy2 + 2y.exy2 = 2xy.y2exy2 + 2y.exy2
= 2xy3exy2 + 2yexy2 = 2xy3exy2 + 2yexy2
(∂ M∂ y
)x = 2xy3exy2 + 2yexy2 (∂N∂ x
)y = 2xy3exy2 + 2yexy2
(∂ M∂ y
)x = (∂N∂ x
)y
(∂F∂ x
)y = M(x,y) (∂F∂ y
) = N(x,y)
∫ dF = ∫M (x,y)dx + h(y)
∫ dF = ∫¿¿2exy2 + 4x3)dx + h(y)
F(x,y) = y2.1/y2exy2 + x4 + h(y)
Mencari nilai h(y) dengan cara mendiferensialkan secara parsial persamaan tersebut
terhadap y...
[∂F∂ y
]x = 2xyexy2 + dh( y )
dy
N(x,y) = 2xyexy2 + dh( y )
dy
2xyexy2 – 3y2 = 2xyexy + dh( y )
dy
– 3y2 = dh( y )
dy
dh(y) = -3y2dy
h(y) = ∫−3y2dy
h(y) = -3.1/3y3
h(y) = -y3
maka hasilnya adalah..
F(x,y) = exy2 + x4 + h(y)
F(x,y) = exy2 + x4 – y3
Pembuktian
Dengan mendiferensialkan hasil secara parsial terhadap x dan y....
F(x,y) = exy2 + x4 – y3
[∂F∂ x
]y = y2exy2 + 4x3
[∂F∂ y
]x = 2xexy2 – 3y2
dF(x,y) = [∂F∂ x
]y dx + [∂F∂ y
]x dy = 0
dF(x,y) = (y2exy2 + 4x3)dx + (2xyexy2 – 3y2)dy
PERSAMAAN DIFFERENSIAL TIDAK EKSAK
Bentuk Umum Persamaan Differensial Tidak Eksak:
m ( x , y ) dx+N ( x , y ) dy=0
( ∂m∂ y )x ≠( ∂ N
∂ x ) y
( ∂m∂ y )x−( ∂ N
∂ x ) y≠ 0
Maka untuk menyelesaikannya harus diubah terlebih dahulu menjadi Persamaan Differensial Eksak dengan cara mengalikannya dengan faktor integral dari persamaan tersebut.
Beberapa faktor integral diantaranya:
Jika:
1. ( ∂m∂ y
−∂ N∂ x
N )=f ( x ) ,maka faktor integralnya fi=e∫ f (x)dx
2. ( ∂m∂ y
−∂ N∂ x
m )=f ( y ) ,maka faktor integralnya fi=e−∫ f ( y)dy
3. mx+Ny ≠ 0 ,maka faktor integralnya fi= 1mx+Ny
4. y f ( x , y ) dx+x g ( x , y )dy=0
f ( x , y )≠ g ( x , y ) ,maka faktor integralnya fi= 1mx−Ny
5. dll
CONTOH :
1. (x¿¿2+ y2+x)dx+xy dy=0¿
Jawab :
(x¿¿2+ y2+x)dx+xy dy=0¿
m ( x , y )= x2+ y2+x N ( x , y )= xy
(∂ M∂ y
¿ x = 2 y (∂N∂ x
¿ y = y
∂ M∂ y
≠∂ N∂ x
persamaan diferensial tidak eksaks.
Maka harus mencari faktor integralnya dulu:
∂ M∂ y
−∂N∂ x
N=
2 y− yxy
¿ yxy
¿1x f ( x )
fi¿e∫ 1
xdx
= e ln x
fi = x
Untuk dijadikan persamaan eksaks, maka dikalikan dengan faktor
integralnya :
x [(x¿¿2+ y2+x)dx+xy dy¿=0
( x3+x y2+x2 ) dx+ x2 y dy=0
m ( x , y )= x3+x y2+x2 N ( x , y ) = x2 y
( ∂ M∂ y )x=2xy ( ∂ N
∂ x ) y=2 xy
∂ M∂ y
=∂ N∂ x
eksaks.
(∂F∂ x
¿ y = m ( x , y ) ; (∂F∂ y
¿ x = N ( x , y )
dF=m ( x , y ) dx
∫ df =∫ ( x3+x y2+x2 ) dx+h( y )
F ( x,y ) = 14
x4+ 12
x2 y2+ 13
x3+h( y )
Selanjutnya mencari nilai h(y) dengan cara mendifferensialkan fungsi
diatas secara parsial terhadap y:
(∂F∂ y
¿ x = x2 y + dh( y )
dy
x2 y = x2 y + dh( y )
dy
dh( y ) = 0 dy
h ( y ) = 0
Maka hasilnya menjadi:
F ( x,y ) = 14
x4+ 12
x2 y2+ 13
x3+h( y )
= 14
x4+ 12
x2 y2+ 13
x3+0
= 14
x4+ 12
x2 y2+ 13
x3
Pembuktian :
Dibuktikan dengan cara mendifferensialkan hasilnya secara parsial
terhadap x maupun y:
14
x4+ 12
x2 y2+ 13
x3 = F ( x,y )
(∂F∂ x
¿ y = ( x3 + xy2 + x2 ) ; (∂F∂ y
¿ x = x2y
d F ( x , y )=( ∂ F∂x ) y dx+( ∂ F
∂ y ) xdy=0
= ( x 3 + xy 2 ) dx + x 2 y dy = 0 : x
= (x2 + y2 + x ) dx + xy dy = 0 terbukti
Soal :
1. (2 x y 4 . e y+2 x y3+ y ) dx+( x2 y 4 .e y−x2 y2−3 x ) dy=0
f. Integral = f(y)
2. (2 x3 y2+4 x2 y+2 x y2+x y 4+2 y ) dx+2 ( y3+x2 y+x ) dy=0 f. Integral = f(x)
3. ( x4+ y 4 ) dx−x y3dy=0 f. Integral = 1
mx+Ny
Jawab :
1. (2 x y 4 . e y+2 x y3+ y ) dx+( x2 y 4 .e y−x2 y2−3 x ) dy=0
m ( x , y )=2 x y 4 .e y+2 x y3+ y
N ( x , y )=x2 y4 . e y−x2 y2−3 x
( ∂m∂ y )x=8 x y3 . ey+e y 2x y4+6 x y2+1
( ∂ N∂ x ) y=2 x y 4 . e y−2 x y2−3
( ∂m∂ y )x ≠( ∂ N
∂ x ) y
maka harus mencari fi nya:
∂m∂ y
−∂N∂ x
m=
8 x y3 . e y+ey 2x y4+6 x y2+1−(2x y4 . e y−2x y2−3)2 x y4 . ey+2 x y3+ y
¿ 8 x y3 e y+8 x y2+42x y4 e y+2x y3+ y
¿4(2 x y3 . e y+2x y2+1)y (2x y3 .e y+2 x y2+1)
¿ 4y
¿4 y−1
maka fi=e−∫ f ( y ) dy
¿e−∫ 4 y−1 dy
¿e−4 ln y
¿e ln y−4
fi= 1
y4
maka :
1
y4 [(2 x y 4 . e y+2 x y3+ y ) dx+( x2 y 4 .e y−x2 y2−3 x ) dy=0 ]
¿ (2 xe y+2 x y−1+ y−3 ) dx+( x2 e y−x2 y−2−3 x y− 4 ) dy=0
m ( x , y )=2 x ey+2 x y−1+ y−3
N ( x , y )=x2 e y−x2 y−2−3x y−4
( ∂m∂ y )x=2x e y−2 x y−2−3 y−4
( ∂ N∂ x ) y=2 xe y−2 x y−2−3 y−4
( ∂m∂ y )x=( ∂ N
∂ x ) y pers. Differensial eksaks.
(∂F∂ x
¿ y = m ( x , y ) ; (∂F∂ y
¿ x = N ( x , y )
( ∂ F∂ x ) y=2x e y+2x y−1+ y−3
dF=2 x e y+2 x y−1+ y−3 dx
∫ dF=∫ (2x e y+2x y−1+ y−3 ) dx+h( y)
F ( x , y )=x2e y+x2 y−1+ y−3+h( y )
persamaan inididifferensialkan secara parsial terhadap y
( ∂ F∂ y )x=x2 e y+(−x2 y−2 )−3x y−4+
d h( y)dy
N ( x , y )=x2 e y−x2 y−2−3xy−4+d h( y )
dy
x2 ey−x2 y−2−3 x y−4= x2 e y−x2 y−2−3 x y−4+dh( y)
dy
0=dh ( y)
dy
h ( y )=0
maka persamaannya menjadi :
F ( x , y )=x2e y+x2 y−1+ y−3+h( y )
¿ x2e y+x2 y−1+ y−3+0
¿ x2e y+x2 y−1+ y−3
Pembuktian :
Dengan cara mendifferensialkan hasil diatas secara parsial terhadap x
maupun terhadap y:
F ( x , y )=x2e y+x2 y−1+ y−3
( ∂ F∂ x ) y=2x e y+2x y−1+ y−3
( ∂ F∂ y )x=x2 e y−x2 y−2−3x y−4
d F ( x , y )=( ∂ F∂x ) y dx+( ∂ F
∂ y ) xdy=0
¿ (2 xe y+2 x y−1+ y−3 ) dx+( x2 e y−x2 y−2−3 x y−4 ) dy=0
terbukti
2. (2 x3 y2+4 x2 y+2 x y2+x y 4+2 y ) dx+2 ( y3+x2 y+x ) d y=0
m ( x , y )=2 x3 y2+4 x2 y+2 x y2+x y4+2 y
N ( x , y )=2 y3+2 x2 y+2 x
( ∂m∂ y )x=4 x3 y+4 x2+4 xy+4 x y3+2
( ∂ N∂ x ) y=4 xy+2
( ∂m∂ y )x ≠( ∂ N
∂ x ) y pers. Diff. Tidak eksak
maka harus mencari fi nya:
∂m∂ y
−∂N∂ x
m=
4 x3 y+4 x2+4 xy+4 x y3+2−(4 xy+2)2 y3+2 x2 y+2 x
¿ 4 x3 y+4 x2+4 x y3
2 y3+2 x2 y+2 x
¿2x (2 y3+2x2 y+2x )(2 y3+2x2 y+2x )
f ( x )=2x
maka fi=e∫f ( x ) dx
¿e∫2x dx
fi=ex2
maka persamaan awaldikalikan dengan fi nya :
ex2
[ (2 x3 y2+4 x2 y+2x y2+x y4+2 y ) dx+2 ( y3+x2 y+x ) dy=0]
¿ (2 x3 y2 ex2
+4 x2 y ex2
+2 x y2ex2
+x y 4 ex2
+2 y ex2 )dx+(2 y3 ex2
+2 x2 y ex2
+2 xex2 )dy=0
m ( x , y )=2 x3 y2 ex2
+4 x2 y ex2
+2 x y2 ex2
+x y 4 ex2
+2 y ex2
N ( x , y )=2 y3 ex2
+2x2 y ex2
+2 x ex2
( ∂m∂ y )x=4 x3 y ex2
+4 x2e x2
+4 xy ex2
+4 x y3ex2
+2ex2
( ∂ N∂ x ) y=2 x .2 y3 ex2
+4 xy ex2
+2xe x2
2x2 y+2ex2
+2 xex2
2 x
¿4 x y3 ex2
+4 xy ex2
+4 x3 y ex2
+2e x2
+4 x2ex2
( ∂m∂ y )x=( ∂ N
∂ x ) y pers. Differensial eksaks.
u v u v
( ∂ F∂ x ) y=m(x , y ) ; ( ∂ F
∂ y )x=N ( x , y )
( ∂ F∂ y )x=2 y3 ex2
+2x2 ye x2
+2 xe x2
¿ (2 y3+2x2 y+2x ) . ex2
dF=( 2 y3+2 x2 y+2 x ) .e x2
dy
∫ dF=∫ (2 y3+2x2 y+2 x ) . ex2
dy+h (x)
F ( x , y )=( 12
y4
+x2 y2+2 xy) . ex2
+h(x )
Lalu mencari nilai h(x) dengan cara mendifferensialkan persamaan
diatas secara parsial terhadap x :
( ∂ F∂ x ) y=(2x y2+2 y ). ex2
+2 xex2( 12
y4
+ x2 y2+2xy )+ dh(x )dx
m ( x , y )=2 x y2 ex2
+2 ye x2
+x y4 ex2
+2x3 y2e x2
+4 x2 y ex2
+dh(x)
dx
2 x3 y2 ex2
+4 x2 yex2
+2 x y2 ex2
+ x y4 e x2
+2 y ex2
=2x3 y2 ex2
+4 x2 y ex2
+2x y2 ex2
+x y4 ex2
+2 y ex2
+dh(x )
dx
0=dh(x)
dx
h(x) = 0
maka hasilnya adalah :
F ( x , y )=( 12
y4
+x2 y2+2 xy) . ex2
+h(x )
¿( 12
y4
+x2 y2+2 xy) . ex2
+0
F ( x , y )=( 12
y4
+x2 y2+2 xy) . ex2
Pembuktian :
u v
F ( x , y )=( 12
y4
+x2 y2+2 xy) . ex2
( ∂ F∂ x ) y=(2x y2+2 y ). ex2
+2 xex2( 12
y4
+ x2 y2+2xy ) ¿2 x y2 ex2
+2 y ex2
+x y 4 ex2
+2 x3 y2 ex2
+4 x2 y ex2
( ∂ F∂ y )x=( 2 y3+2 x2 y+2 x ). ex2
¿2 y3 ex2
+2 x2 y ex2
+2x ex2
dF ( x , y )=( ∂F∂ x ) y dx+( ∂ F
∂ y )x dy=0
¿ (2 x y2 ex2
+2 ye x2
+x y4 ex2
+2x3 y2e x2
+4 x2 y ex2 )dx+(2 y3 ex2
+2 x2 y ex2
+2 xex2 )dy=0
terbukti.
3. ( x4+ y 4 ) dx−x y3dy=0
m ( x , y )=x4+ y4 ; N ( x , y )=− x y3
( ∂m∂ y )x=4 y3 ; ( ∂ N
∂ x ) y=− y3
( ∂m∂ y )x ≠( ∂ N
∂ x ) y pers.diff. tidak eksak
maka harus mencari fi nya:
fi= 1mx+Ny
¿ 1
( x4+ y4 ) x+(−x y3 ) y
¿ 1
x5+x y4−x y4
fi= 1
x5
u
v
selanjutnya persamaanawal dikalikandengan fi nya :
1
x5[ ( x4+ y4 ) dx−x y3 dy=0]
↔ ( x−1+ x−5 y4 ) dx−x−4 y3 dy=0
m ( x , y )=x−1+x−5 y4 ; N ( x , y )=− x−4 y3
( ∂m∂ y )x=( ∂ N
∂ x ) y pers. Differensial eksaks.
( ∂ F∂ x ) y=m(x , y ) ; ( ∂ F
∂ y )x=N ( x , y )
( ∂ F∂ x ) y=x−1+x−5 y4
dF=x−1+x−5 y4 dx
∫ dF=∫(x−1+x−5 y4)dx+h( y )
F ( x , y )=ln x−14
x−4 y4+h( y )
Lalu mencari h(y) dengan cara mendifferensialkan persamaan
tersebut secara parsial terhadap y :
( ∂ F∂ y )x=−x−4 y3+
d h( y )dy
−x−4 y3=− x−4 y3+d h( y)
dy
0=dh ( y)
dy
h ( y )=0
maka hasilnya adalah :
F ( x , y )=ln x−14
x−4 y4+h( y )
¿ ln x−14
x−4 y4+0
F ( x , y )=ln x−14
x−4 y4
Pembuktian :
Dengan cara mendifferensialkan hasil tersebut secara parsial terhadap x
maupun y:
F ( x , y )=ln x−14
x−4 y4
( ∂ F∂ x ) y=1
x+x−5 y4
( ∂ F∂ y )x=−x−4 y3
d F ( x , y )=( ∂ F∂x ) y dx+( ∂ F
∂ y ) xdy=0
¿(x¿¿−1+x−5 y4)dx+ (−x−4 y3 ) dy=0¿
¿(x¿¿−1+x−5 y4)dx−x−4 y3 dy=0¿ terbukti.
PERSAMAAN DIFERENSIAL LINIER TINGKAT n
podn ydxn + p1
dn−1 ydxn−1 + p2
dn−2 ydxn−2 +…+ pn−1
dydx
+ pn y=p ( x )
Penyelesaian persamaan diferensial ada 2:1. Penyelesaian komplementer (yc)
2. Penyelesaian khusus (yp)
Sehingga y= yc+ y p
Penyelesaian komplementer dapat dicari dengan operator (D), D= ddx
[ po Dn+ p1 Dn−1+ p2 Dn−2+…+ pn−1 D+ pn ] y=p(x )
yc=( p0 D n+ p1 Dn−1+p2 Dn−2+…+ pn−1 D+ pn=0)( D−m1) ( D−m2 ) ( D−m3 ) … ( D−mn )=0
Akar-akar persamaan m1, m2, m3, . . . , mn
yc=c1 em1 x+c2em2 x+c3 em3 x+…+cn emn x
Penyelesaian khusus, dari persamaan:
( D−m1) ( D−m2 ) ( D−m3 ) … ( D−mn )=Q(x)
y p=Q(x )
( D−m1) ( D−m2 ) ( D−m3 ) … ( D−mn )
misal :U=Q (x )
( D−m1 )( D−m1) U=Q (x )dudx
−m1U=Q ( x )
U e∫m1 xdx=Q ( x ) e∫m1 xdx
dx
U=U (x )
y p=U ( x )
( D−m2) ( D−m3 ) … ( D−mn )
misal : Z=U ( x )
( D−m2 )( D−m2) Z=U ( x )
dzdx
m2 Z=U ( x )
Ze∫−m2 dx=U ( x ) e∫−m2 dx
dx
Z=Z ( x )
y p
Z (x )( D−m3 ) … ( D−mn )
Contoh:
1.d2 ydx2 −3
dydx
+2 y=ex
[ D2−3 D+2 ] y=ex
( D−2 ) ( D−1 ) y=ex
( D−2 ) ( D−1 )=0m1=2m2=1
yc=c1 e2x+c2ex
y p=ex
( D−2 ) ( D−1 )
misal :U= ex
( D−2 )( D−2 )U=eX
dudx
−2U=ex
U e∫−2 xdx=∫ ex e∫
−2xdxdx
U e−2x=∫ exe−2x dx
U e−2x=∫ e−x dx
U E−2 X=−e−X
U=−ex
sehingga, y p=−ex
( D−1 )
misal : Z= −e x
(D−1 )( D−1 ) Z=−ex
dzdx
−Z=−ex
Ze∫−dx=∫−exe∫
−dxdx
Ze− x=−∫ex e−x dx
Ze− x=−∫dx
Ze− x=−x
Z=−x ex
sehingga, y p−x ex
maka : y= yc+ y p
y=c1 e2x+c2 ex−xex
Pembuktian:y = yc + yp
y=c1 e2x+c2 ex−xex
dydx
=2c1 e2x+c2 ex−[ xe x+e x ]
d2 ydx2 =4c1 e2 x+c2 ex−[ x ex+ex ]−ex
maka :d2 ydx2 −3
dydx
+2 y=ex
4 c1e2x+c2 ex−x ex−ex−e x−3 [2c1 e2x+c2e
x− xex−ex ]+2 [ c1e2x+c2 ex−xe x]=ex
4 c1e2x+c2 ex−x ex−2ex−6c1 e2 x−3c2e
x+3 x ex+3e x+2c1 e2 x+2c2 ex−2x ex=ex
−2ex+3ex=e x
ex=ex
Jikadijumpai :
( D−m1) ( D−m1 ) ( D−m1 ) ( D−m2 )=0
D=m1=m1=m1
yc=( C1+C2+C3 ) em1x+C4 em2x
Jikadijumpai :
D2=−1D=±√−1=± i
yc=C1 eix+C2 e−ix
e ix=cos x+ isin x
e−ix=cos x−i sin x
yc=C1¿¿ (C1+C2 ) cos x+(C1−C2) i sin x
A B¿ A cos x+Bi sin x
Soal :
1.d3 ydx3 +3
d2 ydx2 −4 y=xe−2 x
Jawaban :
d3 ydx3 +3
d2 ydx2 −4 y=x e−2x
[ d3
dx3 +3d2
dx2 −4] y=x e−2x
[ D3+3 D2−4 ] y=x e−2x
Akar−akarnya :1 3 0 -4
2 2 2 -41 1 -2 0
2 2 -21 -1 0
-1 -11 0
( D−1 ) ( D+2 ) ( D+2 ) y=x e−2x
( D−1 ) ( D+2 ) ( D+2 )=0
m1=1 ,m2=−2 ,m3=−2
yc=C1 ex+( C2+C3 ) e−2x
( D−1 ) ( D+2 ) ( D+2 ) y=x e−2x
yp= x e−2x
( D−1 ) ( D+2 ) (D+2 )
u= x e−2x
( D−1 )
( D−1 )u=xe−2 x
dudx
- u = x e−2x→u.e∫−dx=∫ xe−2x . e∫
−dxdx
ue− x=∫ x e−2x . e−x dx
ue− x=∫ x e−3x dx
u dv
dv=e−3x dx
v=−13
e−3x
ue− x=−13
x e−3x —13
e−3 xdx
¿−13
xe−3 x−19
e−3x
MATRIKS
Matriks adalah bilangan atau fungsi yang penulisannya dipisahkan garis
A = 14 B = ex 2 x2
23 sin x cos x
Sub matriks adalah bagian dari matriks induk yang bisa berbentuk kolom atau
mendatar
Sub matriks dari A = [a,j], C = [i,4], D = 4
3
Operasi Matriks
1. Penjumlahan
A = 1 4 , B= 6 2
2 3 1 -5
A + B = 1 4 + 6 2 = 7 6
2 3 1 -5 3 -2
A + B = B + A
2. Pengurangan
A – B= 1 4 - 6 2 = -5 2
2 3 1 -5 1 8
A + B ≠ B + A
3. Perkalian
A . B = 1 4 . 6 2
2 3 1 -5
A . B ≠ B . A
A . C = 1 4 . 2
2 3 4
= 1.2 + 4.4
2.2 + 3.4
= 18
16
A = m x n
B = n x r
A . B = m x r
Metode Transpose
A = 1 4 , AT= 1 2 AT = A
2 3 4 3
Mencari AT dari A
A = 3 2 1 , C = 12 6 -16 , CT = 12 4 12
1 6 3 4 2 16 6 2 -10
2 -4 0 12 10 16 -16 16 1 6
A . CT = 3 2 -1 . 12 4 12
1 6 3 6 2 -10
2 -4 0 -16 16 16
= 64 0 0
0 64 0
0 0 64
A-1 = CT / det A
MATRIKS IDENTITAS
I3 = 1 0 00 1 00 0 1
HCl H2 Cl2
H 1 2 0
Cl 1 0 2
0 0 1
1 2 01 0 20 0 1
1 0 00 1 00 0 1
I A-1
1 0 00 1 00 0 1
1 2 01 0 20 0 1
1 0 00 1 00 0 1
+
1 2 00 −2 20 0 1
1 0 0
−1 1 00 0 1
+
1 2 00 −2 00 0 1
1 0 0
−1 1 −20 0 1
X1
1 0 00 −2 00 0 1
1 1 −2
−1 1 −20 0 1
: -2
I A-1
1 0 00 1 00 0 1
1 1 −212
−12
1
0 0 1
A-1 =
1 1 −212
−12
1
0 0 1
X(-1)
X(-2)
+
A.A-1 = I Cara membuktikan
-2HCl + 1H2 + 1Cl2 = 0
H2 + Cl2 2HCl
PEMBUKTIAN
A . A-1 = I
1 2 01 0 20 0 1
1 1 −212
−12
1
0 0 1
1.0+2.1
2+0.01 .−1 .0 −2+2 .0
0+0+0 1+0+0 −2+0+20+0+0 0+0+0 0+0+1
= 1 0 00 1 00 0 1
= I
Cl2 HCl H2
Cl 1 2 0
H 1 0 2
0 0 1
A I
2 1 00 1 20 0 1
1 0 00 1 00 0 1
+
2 1 00 1 20 0 1
1 0 00 1 −20 0 1
x (-1)
2 0 00 1 00 0 1
1 0 20 1 −20 0 1
I A-1
1 0 00 1 00 0 1
12
0 1
0 1 −20 0 1
X(-2)
+
: 2
Jika : Fe3+ e Fe2+ Fe 1 0 1e 0 1 0 Muatan 3 -1 2
Tidak perlu ditambah I karena sudah 3 baris
A-1 =
12
0 1
0 1 −20 0 1
1Cl2 - 2 HCl + H2 = 0
Cl2 + H2 2 HCl
SOAL
1. N2O5 O2 N2
N 2 0 2O 5 2 0
2. O2 N2O5 N2
N 2 5 2O 0 2 0
JAWABANI A-1
1.
2 0 01 −2 00 0 1
1 0 00 −2 00 0 1
+
X(-2)
X(-2)(-)
: 2(-)
: 2
: 2
A . A-1 = I
=
-1 N2O5 + 5/2 O2 + 1N2 = 0
5/2 O2 + 1N2 N2O5
2. A = A-1
Bukti : A . A-1 = I
=
5/2 O2 – 1 N2O5 + 1N2 = 0
5/2 O2 + 1N2 N2O5
X(-2)+
: 2
(-)
X 5
: 2
Determinat
|A| = det (A) = ∑i
j
aijc ij
a ij = elemen matrik baris ke i kolom ke j
c ij = cofaktor dari elemen matrik baris ke i kolom ke j
c ij = (−1 )i+ j M ij
M ij = minor matrik baris ke i kolom ke j
A = [1 23 4] = [a b
c d ]|A| = 1 . 4 – 2 . 3
= -2
|A| = (−1)1+1.1 . (4 )+(−1)1+2 .2.(3)
= (−1 )2 . ( 4 )+(−1 )3 . (6 )
= 4 – 6
= -2
|A| = (−1)1+2 .2 . (3) + (−1)2+2 . 4 . 1
= -1 . 6 + 4
= -2
A = [1 2 34 5 67 8 9]
|A| = (−1)1+1.1 .|5 68 9|+(−1)1+2 .2 .|4 6
7 9|+(−1)1+3 .3 .|4 57 8|
= 1 .1 (-3) + 2 . 6 + 3 . (-3)
= -3 + 12 + (-9)
= 0
A = [3 2 −11 6 32 −4 0 ]
C = [C11 C12 C13
C21 C22 C23
C31 C32 C33]
C ij = (−1)i+ j M ij
C11 = (−1)1+1 | 6 3−4 0|
= 1 . 12 = 12
C12 = (−1)1+2 |1 32 0|
= (−1)3 . (-6)
= (-1) . (-6)
= 6
C13 = (−1)1+3 |1 62 −4|
= 1 . (-4 - 12)
= -16
C21 = (−1)2+1 | 2 −1−4 0 |
= -1 . (-4)
= 4
C22 = (−1)2+2 |3 −12 0 |
= 1 . (0 – (-2))
= 2
C23 = (−1)2+3 |3 22 −4|
= -1 . (-12 - 4)
= -1 . -16
= 16
C31 = (−1)3+1 |2 −16 3 |
= 1 . (6 – (-6))
= 1 . 12
= 12
C32 = (−1)3+2 |3 −11 3 |
= -1 . (9 – (-1))
= -1 . 10
= -10
C33 = (−1)3+3 |3 21 6|
= 1 . (18 - 2)
= 16
C = [C11 C12 C13
C21 C22 C23
C31 C32 C33] = [12 6 −16
4 2 1612 −10 16 ]
CT = [ 12 4 126 2 −10
−16 16 16 ]A . CT= [3 2 −1
1 6 32 −4 0 ] [ 12 4 12
6 2 −10−16 16 16 ]
= [64 0 00 64 00 0 64]
|A| = (−1)1+1.3 .| 6 3−4 0|+(−1)1+2 .2 .|1 3
2 0|+(−1)1+3 .−1.|1 62 −4|
= 1 . 3 . (0-(-12)) + -1 . 2 . (0 - 6) + 1 . -1 .(-4 - 12)
= 36 + 12 + 16
= 64
A . CT = [det (A ) 0 00 det ( A) 00 0 det (A )]
A . CT = det ( A) [1 0 00 1 00 0 1 ]
A . CT = det ( A) . I
A . CT = det ( A) . A . A−1
CT = det ( A) . A−1
A−1 = CT
det (A )
A−1 = adj( A)det ( A)
Soal :
Penerapan dalam kimia
log ¿Io
=ECtx=A
λ1 = 3x + 2y + 5z = 9
λ2 = x + 3y + 2z = 18
λ3 = 2x + y + 4z = 12
A = [3 2 51 3 22 1 4 ] B = [ xyz ] C = [ 9
812]
|A| = (-1)1+1 3 |3 21 4| + (-1)1+2 (2) |1 2
2 4| + (-1)1+3|3 32 1| = 5
A-1 = ? C = [ 10 0 −5−3 2 1−11 −1 7 ]
-logI t
I0 = ∑C t = A
λ1 => 3x + 2y + 5z = 9
x + 3y + 2z = 8 2x + y + 4z = 12
I0 It
UV
x,y,z
λ
A
λ1
y
λ
A
λ2
y
λ
A
λ3
y
A = [3 2 51 3 22 1 4 ] B = [ xyz ] C = [ 9
812]
C = [C11 C12 C13
C21 C22 C23
C31 C32 C33]
C11 = (−1)1+1 |3 21 4|
= 1 . (12 – 2)
= 10
C12 = (−1)1+2 |1 22 4|
= -1 . (4 - 4)
= (-1) . 0
= 0
C13 = (−1)1+3 |1 32 1|
= 1 . (1 - 6)
= -5
C21 = (−1)2+1 |2 51 4|
= -1 . (8 - 5)
= -3
C22 = (−1)2+2 |3 52 4|
= 1 . (12 – 10)
= 2
C23 = (−1)2+3 |3 22 1|
= -1 . (3 - 4)
= -1 . -1
= 1
C31 = (−1)3+1 |2 53 2|
= 1 . (4 – 15)
= 1 . -11
= -11
C32 = (−1)3+2 |3 51 2|
= -1 . (6 – 5)
= -1. 1
= -1
C33 = (−1)3+3 |3 21 3|
= 1 . (9 - 2)
= 7
C = [C11 C12 C13
C21 C22 C23
C31 C32 C33] = [ 10 0 −5
−3 2 1−11 −1 7 ]
CT = [ 10 −3 −110 2 −1
−5 1 7 ] A−1 =
15 [ 10 −3 −11
0 2 −1−5 1 7 ]
A−1 . D=[ 2−35
−115
025
−15
−115
75
] [ 9812]
A−1 . D=[ 18−24
5−11
5
0+16
5−12
5
−9+85
+845
]A−1 . D=[
−66545475
]Pembuktian :
λ1=3 x+2 y+5 z=9
λ1=3(−665 )+2
45+5
475
=9
λ1=−198
5+ 8
5+235
5=9
λ1=9=9 terbukti
λ2=x+3 y+2 z=8
λ2=(−665 )+3
45+2
475
=8
λ2=−66
5+ 12
5+ 94
5=8
λ2=8=8 terbukti
λ3=2x+ y+4 z=12
λ3=2(−665 )+ 4
5+4
475
=12
λ3=−132
5+ 4
5+ 188
5=12
λ3=12=12 terbukti
B=[ xyz ]=[−66
545475
]A1=[ 9 2 5
8 3 212 1 4 ]
|A1|=(−1 )2 .9|3 21 4|+(−1 )3 2| 8 2
12 4|+ (−1 )4 5| 8 312 1|
|A1|=9 (12−2 )+(−2 ) (32−24 )+5 (8−36 )
|A1|=90−16−140=−66
A2=[3 9 51 8 22 12 4 ]
|A2|=(−1 )2 .3| 8 212 4|+(−1 )3 9|1 2
2 4|+ (−1 )4 5|1 82 12|
|A2|=3 (32−24 )+(−9 ) (4−4 )+5 (12−16 )
|A2|=24+0−20=4
A3=[3 2 91 3 82 1 12]
|A3|=(−1 )2 .3|3 81 12|+(−1 )3 2|1 8
2 12|+(−1 ) 4 9|1 32 1|
|A3|=3 (36−8 )+(−2 ) (12−16 )+9(1−6)
|A3|=84+8−45=47
A−1 . D=[ xyz ]=[|A1||A||A2||A||A3||A|
]
C = CΠ
σ jiH
H
H
H
C = CΠ
σ jiH
H
H
H
Terapan dalam ikatan kimia
H ij=∫ψ1 Ĥ ψ jdτ
apabila i = j α
apabila i ≠ j (ada ikatan langsung) β
apabila i ≠ j (tidak ada ikatan langsung) 0
Sij=∫ψ1ψ j dτ
Jika i = j 1
Jika i ≠ j 0 , [terhadap dirinya sendiri (1,1) atau (2,2 ),dll]
H ij−E .S ij=0
C11 ;α−E .1=0
C12; β−E .0=0
C21; β−E .0=0
C22;α−E .1=0
[α−E ββ α−E]=0 dibagi dengan β
[ α−Eβ
1
1α−E
β]=0
Misal α−E
β=x
E
α
α+β
α -β
C = C – C = C
C = C – C = C
[ x 11 x ]=0
|x 11 x|=x2−1=0
X2 = 1
X1 = -1
X2 = 1
α−Eβ
=−1
α−E=−β
E=α+ β
α−Eβ
=1
α−E=β
E=α−β
Nilai β ˂ 0
Gambar :
Soal
1. 1,3 butadiena
2. Benzena
Jawaban
1.
C11 ;α−E .1=0
C12; β−E .0=0
C13;0−E .0=0
C14;0−E .0=0
C21; β−E .0=0
C22;α−E .1=0
C23; β−E .0=0
C24;0−E .0=0
C31;0−E .0=0
C32; β−E .0=0
C33;α−E .1=0
C34; β−E .0=0
C41;0−E .0=0
C42;0−E .0=0
C43; β−E .0=0
C44 ;α−E .1=0
[α−E β 0 0β α−E β 000
β0
α−Eβ
βα−E
]=0
[α−E
β1 0 0
1α−E
β1 0
00
10
α−Eβ1
1α−E
β]=0
Misal : α−E
β=x
[ x 1 0 01 x 1 000
10
x1
1x]=0
|x 1 0 01 x 1 000
10
x1
1x|=x4−3x2+1
x=±( 12
(1+√5 ))=±1,618
x=±( 12
(√5−1 ))=± 0,618
α - 1,618 β
α - 0,618 β
α + 0,618 β
α + 1,618 β
E
α
1.α−E
β=1,618
α−E=1,618 βE=α−1,618 β
2.α−E
β=−1,618
α−E=−1,618 βE=α+1,618 β
3.α−E
β=0,618
α−E=0,618 βE=α−0,618 β
4.α−E
β=−0,618
α−E=−0,618 β
E=α+0,618 β
Gambar :
2.Benzena
C11 ;α−E .1=0
C12; β−E .0=0
C13;0−E .0=0
C14;0−E .0=0
C15;0−E .0=0
C16; β−E .0=0
C21; β−E .0=0
C22;α−E .1=0
C23; β−E .0=0
C24;0−E .0=0
C25;0−E .0=0
C26;0−E .0=0
C31;0−E .0=0
C32; β−E .0=0
C33;α−E .1=0
C34; β−E .0=0
C35;0−E .0=0
C36;0−E .0=0
C41;0−E .0=0
C42;0−E .0=0
C43; β−E .0=0
C44 ;α−E .1=0
C45; β−E .0=0
C46;0−E .0=0
C51;0−E .0=0
C52;0−E .0=0
C53;0−E .0=0
C54; β−E .0=0
C55;α−E .1=0
C56; β−E .0=0
C61; β−E .0=0
C62;0−E .0=0
C63;0−E .0=0
C64;0−E .0=0
C65; β−E .0=0
C66;α−E .1=0
[α−E β 0 0 0 β
β α−E β 0 0 0000β
β000
α−Eβ00
βα−E
β0
0β
α−Eβ
00β
α−E] dibagi β
[α−E
β1 0 0 0 1
1α−E
β1 0 0 0
0001
1000
α−Eβ100
1α−E
β10
01
α−Eβ1
001
α−Eβ
]Misal
α−Eβ
=x
[x 1 0 0 0 11 x 1 0 0 00001
1000
x100
1x10
01x1
001x]
|x 1 0 0 0 11 x 1 0 0 00001
1000
x100
1x10
01x1
001x|=x6−6 x4+9 x2−4
x=± 2x=± 1
1.α−E
β=2
α−E=2 βE=α−2 β
2.α−E
β=−2
α−E=−2 βE=α+2β
3.α−E
β=1
α−E=βE=α−β
4.α−E
β=−1
α−E=−β
E=α+ β
Cahaya Εn=4 ❑α−1,6 β Melepas e
En=3 ❑α−0,6 β
α ------------------------------
En=2 ↑↓α+0,6 β
En=1 ↑↓α+1,6 β