Tugas akhir matematika kelompok 1
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1. Tunjukkan bahwa barisan yang didefinisikan dengan an = 1; an+1 = 3 - 1
ππ adalah
barisan yang increasing dan an < 3 untuk setiap n
Jawab :
Syarat barisan yang increasing : an < an+1
Dik : an = 1
an+1 = 3 - 1
ππ
an < 3
Maka, an < an+1 Ι n β₯ 3
an < 3 - 1
ππ Ι n β₯ 3
1 < 3 - 1
1 Ι n β₯ 3
1 < 2
Terbukti sebagai barisan yang increasing
1 < 2 an < an+1 Ι n β₯ 3
2. Perhatikan sketsa grafik berikut :
Jika f kontinu pada [a,b] dan c β¬ [a,b] demikian sehingga :
f (c) = faverage = 1
πβπ β« π(π₯)ππ₯
π
π
maka, β« π(π₯)ππ₯π
π = f(c) (b-a)
Hitung faverage dari grafik sketsa berikut :
A. Cari nilai faverage
B. Cari nilai b demikian sehingga Nilai rata-rata (average) dari f(x) = 2+6x-3x2 pada
interval [0,b] sama dengan 3.
Jawab :
A. f(c) = 1
πβπ β« π(π₯)ππ₯
π
π
1
2β(β1) β« 1 + π₯22
β1
1
3 [x+
1
3 x3]β1
2
1
3 [
14
3 +
4
3 ]
1
3 [
18
3 ] =
6
3
faverage = 2
B. f(c) = 1
πβπ β« π(π₯)ππ₯
π
π
3 = 1
π β« 2 + 6π₯ β 3π₯2π
0
3 = 1
π [ 2x+3x2-x3 ]0
π
3 = 1
π [ 2b+3b2-b3 ]
3 = 2+3b-b2
-b2 + 3b β 1= 0
b2 β 3b +1 = 0
3. Cari Solusi dari PD : 2xyyβ β y2 + x2 = 0
Jawab :
2xyyβ β y2 + x2 = 0
2xy ππ¦
ππ₯ β y2 + x2 = 0
2xy ππ¦
ππ₯ = y2 - x2
ππ¦
ππ₯ =
π¦2βπ₯2
2π₯π¦
2xy dy = (y2 β x2) dx
2xy = M(x,y) ; y2 β x2 = N(x,y)
Diubah jadi M(ππ₯, Ξ»π¦)
Uji PD homogen :
M(x,y) = M(ππ₯, Ξ»π¦) kita ambil π = 3
32 2xy = 2.3x.3y
32 2xy = 32 2xy
πn N(x,y) = N (ππ₯, Ξ»π¦)
32(y2-x2) = (3y)2 - (3x)2
32(y2-x2) = 9y2-9x2
9y2-9x2 = 9y2-9x2
Terbukti PD Homogen
2xyyβ β y2 + x2 = 0
2xy ππ¦
ππ₯ β y2 + x2 = 0
b1 = 2,618 ; b2 = 0,3819
2xy ππ¦
ππ₯ = y2 - x2
2xy dy = (y2 - x2) dx
(y2 - x2) dx - 2xy dy = 0
M(x,y) dx β N(x,y) dy = 0
Solusi : y = vx
dy = vdx + xdv
(y2 - x2) dx - 2xy dy = 0
[(vx)2 β x2] dx β 2x(vx)(vdx+xdv) = 0
[v2x2 β x2] dx β 2x2v (vdx+xdv) = 0
v2x2dx β x2dx β 2x2v2dx β 2x3 vdv = 0
(v2x2 - x2 β 2x2v2) dx - 2x3 vdv = 0
(-x- v2x2) dx β 2x3 vdv = 0
M (x,v) N (x,v)
-x2 ( 1+v2) dx β 2x3 vdv = 0
βπ₯2(1+π£2)ππ₯
π₯3(1+ π£2) -
2π₯3 π£ππ£
π₯3(1+π£2) = 0
β1
π₯ππ₯ β
2π£
(1+π£2)ππ£ = 0
β β«1
π₯ ππ₯ β β«
2π£
(1+π£2)ππ£ = 0
β β«1
π₯ ππ₯ = β«
2π£
(1+π£2)ππ£
-ln |x| + C1 = ln |1+v2| + C2
-ln |x| - ln | 1+v2 | = C2 - C1
- (ln |x| - ln | 1+v2 |) = C2 - C1
ln |x| + ln | 1+v2 | = -(C2 - C1)
ln |x| + ln | 1+v2 | = C2 + C1
ln |x| + ln | 1+v2 | = ln ππ2+π1
ln |x| | 1+v2 | = ln ππ2+π1
|x| | 1+v2 | = ππ2+π1
x(1+v2) = C
x+xv2= C
solusi y= v.x
v= π¦
π₯
x+ x(π¦
π₯)2 = C
x + π¦2
π₯ = C
y2= ( C-x ) x
y2= Cx β x2
y = βππ₯ β π₯2
Jadi solusi dari PD 2xyyβ β y2 + x2 = 0 adalah y = βππ₯ β π₯2
Catatan :
-β«1
π₯ dx = -ln |x|
β«2π£
(1+π£2) dv
Misal : u = 1+v2
ππ’
ππ£ = 2v
dv = 1
2π£ du
= β«ππ’
π’
= ln |u| + C
= ln |1+v2| + C