Tug As 2
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Ubahlah PD berikut menjadi pers. Bessel dengan substitusinya yang diberikan. Carilah solusi tersebut dan diucapkan dalam fungsi Bessel: 3
2 z = x2 y " + xy = 0 , dengan y = u x , 3 Penyelesaian: 3
3 2
2 z = x2 " y + xy = 0 , y = u x , 3
x =
3
2 z = x2 y = u (x ) 3 Mis, , 1 2
−1
1
dy du 1 2 = x2 + x u dx dx 2 •
−1
1 2
du dz 1 2 =x + x u dz dx 2 1
= x2
−1
1
du 2 1 2 x + x u dz 2 −1
du 1 2 =x + x u dz 2 1 d 2 y d du 1 − 2 = x + x u dx 2 dx dz 2
• −3
−1
=
du d 2u 1 2 1 du +x − x u+ x2 dz dxdz 4 2 dx
=
du d 2 u dz 1 2 1 du dz +x 2 − x u+ x2 dz 2 dz dx dz dx 4
−3
−1
3 z 2
−3
3
=
du d 2u 1 1 du + x2 2 − x 2 u + dz 4 2 dz dz −3
3 2
d 2 u 3 du 1 2 =x + − x u dz 2 2 dz 4
•
y " + xy = 0 32 d 2 u 3 du 1 −23 12 x + − x u + x ux = 0 dz 2 2 dz 4 −3
3 2
3
d 2 u 3 du 1 2 x + − x u + ux 2 = 0 dz 2 2 dz 4 3
x2
3
x2
3 −3 d 2 u 3 du 2 1 2 u = 0 + + x − x 4 dz 2 2 dz
3 d 2 u 3 du 2 1 + + x − 3 2 2 dz dz 4x 2
3 2 u = 0 → x 4 x
2 3 3 2 32 d 2 u 3 du 2 u = 0 2 4 x + . 4 x + 4 x − 1 2 2 dz dz
2 2 2 3 du 3 3 d u u = 0 4 z + 6 . z + 4 z − 1 2 2 2 dz 2 dz
9 d 2u du 9 2 4. z 2 2 + 9 z + 4. z − 1u = 0 4 dz dz 4 9z 2
z2
(
)
d 2u du 1 + 9z + 9z 2 − 1 u = 0 → x 2 dz 9 dz
d 2u du 2 1 +z + z − u = 0 2 dz 9 dz
Pers. Bessel dengan variabel z dan
v=
1 3 , sehingga solusi umum:
u ( x ) = A0 J 1 ( z ( x ) ) + B0 J −1 ( z ( x ) ) 3
3
2 32 2 32 u ( x ) = A0 J 1 x + B0 J −1 x 3 3 3 3
Karena y = u x , jadi: 2 32 2 32 y ( x ) = x A0 J 1 x + B0 J −1 x 3 3 3 3 2 3 1 2 3 1 = A0 J 1 x 2 .x 2 + B0 J −1 x 2 .x 2 3 3 3 3 =
2 2 2 A0 J 1 x + B0 J −1 x 3 3 3
( )
( )
2 z = x2 y " + xy = 0 , dengan y = u x , 3 Penyelesaian: 3
3 2
2 z = x2 " y + xy = 0 , y = u x , 3
x =
3
2 z = x2 y = u (x ) 3 Mis, , 1 2
−1
1
dy du 1 2 = x2 + x u dx dx 2 •
−1
1 2
du dz 1 2 =x + x u dz dx 2 1
= x2
−1
1
du 2 1 2 x + x u dz 2 −1
du 1 2 =x + x u dz 2 1 d 2 y d du 1 − 2 = x + x u dx 2 dx dz 2
• −3
−1
=
du d 2u 1 2 1 du +x − x u+ x2 dz dxdz 4 2 dx
=
du d 2 u dz 1 2 1 du dz +x 2 − x u+ x2 dz 2 dz dx dz dx 4
−3
−1
3 z 2
−3
3
=
du d 2u 1 1 du + x2 2 − x 2 u + dz 4 2 dz dz −3
3 2
d 2 u 3 du 1 2 =x + − x u dz 2 2 dz 4
•
y " + xy = 0 32 d 2 u 3 du 1 −23 12 x + − x u + x ux = 0 dz 2 2 dz 4 −3
3 2
3
d 2 u 3 du 1 2 x + − x u + ux 2 = 0 dz 2 2 dz 4 3
x2
3
x2
3 −3 d 2 u 3 du 2 1 2 u = 0 + + x − x 4 dz 2 2 dz
3 d 2 u 3 du 2 1 + + x − 3 2 2 dz dz 4x 2
3 2 u = 0 → x 4 x
2 3 3 2 32 d 2 u 3 du 2 u = 0 2 4 x + . 4 x + 4 x − 1 2 2 dz dz
2 2 2 3 du 3 3 d u u = 0 4 z + 6 . z + 4 z − 1 2 2 2 dz 2 dz
9 d 2u du 9 2 4. z 2 2 + 9 z + 4. z − 1u = 0 4 dz dz 4 9z 2
z2
(
)
d 2u du 1 + 9z + 9z 2 − 1 u = 0 → x 2 dz 9 dz
d 2u du 2 1 +z + z − u = 0 2 dz 9 dz
Pers. Bessel dengan variabel z dan
v=
1 3 , sehingga solusi umum:
u ( x ) = A0 J 1 ( z ( x ) ) + B0 J −1 ( z ( x ) ) 3
3
2 32 2 32 u ( x ) = A0 J 1 x + B0 J −1 x 3 3 3 3
Karena y = u x , jadi: 2 32 2 32 y ( x ) = x A0 J 1 x + B0 J −1 x 3 3 3 3 2 3 1 2 3 1 = A0 J 1 x 2 .x 2 + B0 J −1 x 2 .x 2 3 3 3 3 =
2 2 2 A0 J 1 x + B0 J −1 x 3 3 3
( )
( )
