1. Tunjukkan bahwa barisan yang didefinisikan dengan an = 1; an+1 = 3 - 1

𝑎𝑛 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

𝑎𝑛

an < 3

Maka, an < an+1 ɏ n ≥ 3

an < 3 - 1

𝑎𝑛 ɏ n ≥ 3

1 < 3 - 1

1 ɏ n ≥ 3

1 < 2

Terbukti sebagai barisan yang increasing

1 < 2 an < an+1 ɏ n ≥ 3

2. Perhatikan sketsa grafik berikut :

Jika f kontinu pada [a,b] dan c € [a,b] demikian sehingga :

f (c) = faverage = 1

𝑏−𝑎 ∫ 𝑓(𝑥)𝑑𝑥

𝑏

𝑎

maka, ∫ 𝑓(𝑥)𝑑𝑥𝑏

𝑎 = 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

𝑏−𝑎 ∫ 𝑓(𝑥)𝑑𝑥

𝑏

𝑎

1

2−(−1) ∫ 1 + 𝑥22

−1

1

3 [x+

1

3 x3]−1

2

1

3 [

14

3 +

4

3 ]

1

3 [

18

3 ] =

6

3

faverage = 2

B. f(c) = 1

𝑏−𝑎 ∫ 𝑓(𝑥)𝑑𝑥

𝑏

𝑎

3 = 1

𝑏 ∫ 2 + 6𝑥 − 3𝑥2𝑏

0

3 = 1

𝑏 [ 2x+3x2-x3 ]0

𝑏

3 = 1

𝑏 [ 2b+3b2-b3 ]

3 = 2+3b-b2

-b2 + 3b – 1= 0

b2 – 3b +1 = 0

3. Cari Solusi dari PD : 2xyy’ – y2 + x2 = 0

Jawab :

2xyy’ – y2 + x2 = 0

2xy 𝑑𝑦

𝑑𝑥 – y2 + x2 = 0

2xy 𝑑𝑦

𝑑𝑥 = y2 - x2

𝑑𝑦

𝑑𝑥 =

𝑦2−𝑥2

2𝑥𝑦

2xy dy = (y2 – x2) dx

2xy = M(x,y) ; y2 – x2 = N(x,y)

Diubah jadi M(𝜆𝑥, λ𝑦)

Uji PD homogen :

M(x,y) = M(𝜆𝑥, λ𝑦) kita ambil 𝜆 = 3

32 2xy = 2.3x.3y

32 2xy = 32 2xy

𝜆n N(x,y) = N (𝜆𝑥, λ𝑦)

32(y2-x2) = (3y)2 - (3x)2

32(y2-x2) = 9y2-9x2

9y2-9x2 = 9y2-9x2

Terbukti PD Homogen

2xyy’ – y2 + x2 = 0

2xy 𝑑𝑦

𝑑𝑥 – y2 + x2 = 0

b1 = 2,618 ; b2 = 0,3819

2xy 𝑑𝑦

𝑑𝑥 = y2 - x2

2xy dy = (y2 - x2) dx

(y2 - x2) dx - 2xy dy = 0

M(x,y) dx – N(x,y) dy = 0

Solusi : y = vx

dy = vdx + xdv

(y2 - x2) dx - 2xy dy = 0

[(vx)2 – x2] dx – 2x(vx)(vdx+xdv) = 0

[v2x2 – x2] dx – 2x2v (vdx+xdv) = 0

v2x2dx – x2dx – 2x2v2dx – 2x3 vdv = 0

(v2x2 - x2 – 2x2v2) dx - 2x3 vdv = 0

(-x- v2x2) dx – 2x3 vdv = 0

M (x,v) N (x,v)

-x2 ( 1+v2) dx – 2x3 vdv = 0

−𝑥2(1+𝑣2)𝑑𝑥

𝑥3(1+ 𝑣2) -

2𝑥3 𝑣𝑑𝑣

𝑥3(1+𝑣2) = 0

−1

𝑥𝑑𝑥 −

2𝑣

(1+𝑣2)𝑑𝑣 = 0

− ∫1

𝑥 𝑑𝑥 − ∫

2𝑣

(1+𝑣2)𝑑𝑣 = 0

− ∫1

𝑥 𝑑𝑥 = ∫

2𝑣

(1+𝑣2)𝑑𝑣

-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 𝑒𝑐2+𝑐1

ln |x| | 1+v2 | = ln 𝑒𝑐2+𝑐1

|x| | 1+v2 | = 𝑒𝑐2+𝑐1

x(1+v2) = C

x+xv2= C

solusi y= v.x

v= 𝑦

𝑥

x+ x(𝑦

𝑥)2 = C

x + 𝑦2

𝑥 = C

y2= ( C-x ) x

y2= Cx – x2

y = √𝑐𝑥 − 𝑥2

Jadi solusi dari PD 2xyy’ – y2 + x2 = 0 adalah y = √𝑐𝑥 − 𝑥2

Catatan :

-∫1

𝑥 dx = -ln |x|

∫2𝑣

(1+𝑣2) dv

Misal : u = 1+v2

𝑑𝑢

𝑑𝑣 = 2v

dv = 1

2𝑣 du

= ∫𝑑𝑢

𝑢

= ln |u| + C

= ln |1+v2| + C