Nama Kelompok: Dwi Kurniati Zaenab (105017000416) Rahmadini Husna (105017000434) Mas’udah (105017000468) Tugas Nilai Awal Syarat Batas Metode Deret y"+ y ' = 0 2a 2 + 6a3 x + 12a 4 x 2 + ....... + a1 + 2a 2 x + 3a3 x 2 + ..... = 0 (2a 2 + a1 ) + (6a3 + 2a 2 ) x + (12a 4 + 3a3 ) x 2 + ........ = 0 2a 2 + aa1 = 0 a1 2 6a 3 + 2a 2 = 0 a2 = −
2a 2 2 a a = − − 1 = 1 6 6 2 6 12a 4 + 3a3 = 0 a3 = −
a4 = −
3a3 a 3 a =− 1=− 1 12 12 6 24
maka : y = a 0 + a1 x + a 2 x 2 + a3 x 3 + a 4 x 4 + .... a1 2 a1 3 a1 4 x + x − x + ..... 2 6 24 x2 x3 x4 = a 0 + a1 x − + − + ...... 2 6 24 = a 0 + a1 x −
[
= a 0 + a1 1 − e − x = a 0 + a1 − a1e
]
−x
= A1 + A2 e − x Misal: a0 + a1 = A1
dan –a1 = A2
Metode Biasa (Integral) y"+ y ' = 0 Misal : d = D sehingga dy = Dy dx dx d2y d2 2 Sehingga = D2 y = D 2 2 dx dx y"+ y ' = 0 d 2 y dy + =0 dx 2 dx D 2 y + Dy = 0 ( D 2 + D) y = 0 D( D + 1) y = 0 D = 0atauD = −1 solusi : y = c1e 0 x + c 2 e − x y = c1 + c 2 e − x