Persamaan Diferensial Orde II
PDB Orde II PDB Orde II
″ ′ ″ ′ Solusi Homogen Solusi Homogen
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# # $ $
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± − 4 . 1 .
4 = ± 2 i
=
r
12
2
( ) " ( "
% " Persamaan Differensial non Persamaan Differensial non homogen homogen
″ ′ ≠
/ !
% ! " !
Metode koefisien tak tentu Metode koefisien tak tentu
r(x) y p r(x) = e mx y p = A e mx r(x) = X n y p = A n
X n + A n-1
X n-1 +…….+A 1 X + A r(x) = sin wx y p = A cos wx + B sin wx r(x) =cos wx y p = A cos wx + B sin wx r(x) = e ux sin wx y p = e ux (A cos wx + B sin wx ) R(x) =e ux cos wx y p = e ux (A cos wx + B sin wx ) ( /
) # " Contoh Contoh
% 1 2 $ 3 " # 4
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Contoh (no. 2 Lanjutan) Contoh (no. 2 Lanjutan)
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Latihan Latihan
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Metode Variasi Parameter Metode Variasi Parameter ′ ′ # ′ ″ ′ ′ ″
′
# ′ ′ $ %
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%
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% # $
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#
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$ # % $ % $
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4
/ ″
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#
′ ′
′ ′ # ′ ′ $ %99999 ::
Metode Variasi Parameter Metode Variasi Parameter
; : ::
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#
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)
y y 2 1 r ( x ) y ' 2 y r ( x ) y ' r ( x ) 2 1 y r ( x ) 1
= = − u ' u dx = = v ' v dx y y 1 2 W y y 1 2 W y ' y ' 1 2 y ' y ' 1 2 y y 1 2
6 W =
y y 1 ' ' 2 Contoh Contoh
% & # 4
"
% < # #
( ( ) ) ( (
' ' % % " "
5 #
) (
" "
3 , ) %
) / 2 2
x − sin 1 cos x x x sin tan
= − x − x dx = − dx = − (sec cos ) dx u = − dx x cos cos x
1 Contoh (Lanjutan) Contoh (Lanjutan) x x x sin tan sec ln
- − = + − = dx x dx x cos sec = dx
x x v 1 tan cos
= dx x sin x cos
− =
/ # #
( ) x x x x x x x y p cos sin cos sin cos tan sec ln
− + + − =
- − =
#
( ) x x x cos tan sec ln
- − + =
( ) C x x x x x C y cos tan sec ln sin cos 2 1
- < $ #
$ /
3
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$ ) $ (
$ )
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$
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− = dx x x u
%
3 3 sec 3 sin 2 − = dx x
3 tan
3
1
2
( )− − = dx x
1 3 sec
3
,$ $
3
Contoh Contoh
(
" & ) # 4
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(
%
) $8 (
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$8 #
%
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5
% " %
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1 2
- =
- − =
3
1 3 cos
3
1
#
( ) C x x x x x x x C y
3 sin 3 tan 3 sec ln
9
1 3 cos
1 3 sin
1 3 cos 3 tan
9
1 3 cos 2 1
( ) x x x x x x
3 sin 3 tan 3 sec ln
9
1 3 sin
9
1 3 cos
3
9
9
1
3
Contoh (Lanjutan) Contoh (Lanjutan) x x
3 tan
9
1
3
1 − = − = dx x dx 3 sec
3
1
1 2 = dx x x v
3 sin 3 tan 3 sec ln
3 3 sec 3 cos 2 = dx x
3 sec
3
1 x x
3 tan 3 sec ln
9
1
/ # #
( ) x x x x x x x y p
- − =
- − + =
- e
1 − 2 x e e 4. y” + 4 y’ + 4 y =
2 x 5. y” + 4 y = 3 cosec 2x
6. y” + 4 y = 3 cosec x 7. 4 y” + y = 2 sec (x/2) x e 8. y” – 2y’ + y =
2
- 1 x
Penerapan dalam Rangkaian Listrik Penerapan dalam Rangkaian Listrik
)
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- ,$-% 0 = 6 ) .
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- ) . , - = + +
( ) "
- /- /-
(Lanjutan) (Lanjutan) /.
&
0 = /-
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/- /- +
Contoh Contoh
? @ A
4 BC( ) '1 * D
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4 4 / )
BC(
= " . E %+ . > . %"
&
- = . E F "* + . > .
Contoh Contoh
"* F " = + + $ . ± − =
2 /
- = −
( ) - + - + . - '
$ $ ) " % .
/
- = − − $
( ) - + - + . -
$ $ ) % . " " % . $
% . "
− = ..
2 #
Rangkaian RLC Rangkaian RLC
- #
- $ %
- − = − −
( )
[ ] - - . -
$ " $ $ ) . " . " % . $ − −
4 .
$ " % " $ − − =
% . " − − =
( )
[ ] - - . -
$ " $ $ ) . " . " % . $
$ " > . − = =
- - . -
-
Latihan Latihan
% = BC( ) '1 *
- ,$-% -
4 " ? @
4 B( )
- , D
4
Latihan Latihan
$ = BC( ) % *
$ *
,+
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4
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= + %
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4