Tug As

Ikfi Mubarokah (106017000522) Mulia Rahmayani (106017000533) Tugas Nilai Awal Syarat Batas Soal persamaan diferensial orde-kedua a) Dengan menggunakan cara biasa : 1. y" + 16 y = 0 y ( y + 16 ) = 0 y = 0 , y = -16 y1 = C1

, y2 = C2

Jadi solusinya: y = y1 + y2 y = C1

+ C2

y = C1 + b) Dengan menggunakan metode deret pangkat : 1. y" + 16 y = 0 misalkan : y = a 0 + a1 x + a2 x2 + a3 x3 + a 4 x4 + a5 x5+ … y" = 2a2 + 6a3 x + 12a4 x2 + 20a5 x3 + 30a6 x4 + 42a7 x5 + … jadi : y" + 16 y = 0 •

( 2a2 + 6a3 x + 12a4 x2 + 20a5 x3 + 30a6 x4 + 42a7 x5 + … ) + 16 ( a 0 + a1 x + a2 x2 + a3

x3 + a 4 x4 + a5 x5+ … ) •

( 2a2 + 6a3 x + 12a4 x2 + 20a5 x3 + 30a6 x4 + 42a7 x5 + … ) + ( 16a 0 + 16a1 x + 16a2 x2 + 16a3 x3 + 16a 4 x4 + 16a5 x5+ … )

•

( 2a2 + 16a 0 ) + ( 6a3 x + 16a1 x ) + ( 12a4 x2 + 16a2 x2 ) + ( 20a5 x3 + 16a3 x3 ) + ( 30a6 x4 + 16a 4 x4 ) + ( 42a7 x5 + 16a5 x5) + . . .

•

( 2a2 + 16a 0 ) + x ( 6a3 + 16a1 ) + x2 ( 12a4 + 16a2 ) + x3 ( 20a5 + 16a3 ) + x4 ( 30a6 + 16a 4 ) + x4 ( 42a7 + 16a5) + . . .

Sehingga diperoleh : 2a2 + 16a 0 = 0 => a2 = 6a3 + 16a1 = 0

= - 8a 0 => a3 =

=

12a4 + 16a2 = 0 => a4 =

=

=

20a5 + 16a3 = 0=> a5 =

=

=

=> a6 =

=

=

=

30a6 + 16a 4 = 0 42a7 + 16a5 = 0=> a7 =

=

Jadi solusinya; y = a0 + a1 x - 8a 0 x2 y = a0 ( 1 -8 x2 + y = a0 ( 1 –

x3 +

x4 +

x4 +

x5 -

x6 -

x7 + …

x6 + … ) + a1 ( x -

x3 +

x5 -

+ … ) + a1 ( x –

+

–

–

y = a0 cos(4x) + 2a1 sin(2x)

x7 + … ) +…)