Question

TOLONG DIBANTU. PAKE CARANYA YAA. MAKASIH SEBELUMNYA

Answer 1

Nomor 10.
y' = x/4y
4y.y' = x
Integralkan.
2y² = x²/2 + C
x²/2 - 2y² - C = 0
Melalui (2,-1)
2 - 2 - C = 0
C = 0
Sehingga, fungsi implisitnya adalah:
F(x,y) = x²/2 - 2y²

Nomor 11.
f'(x) = 1/√x + √x
f'(x) = x^(-½) + x^½
Sehingga,
f(x) = 2.x^½ + 2/3 x^(3/2) + C
Dengan, f(4) = 0
0 = 2(2) + 2/3 (8) + C
C = -4 - 16/3
C = -28/3
Didapat:
f(x) = 2√x + 2/3 x√x - 28/3

Nomor 12.
y' = 1 - 16.x⁻⁴
y = x + 16x⁻³/3 + C
Sehingga, y(1) = 2
2 = 1 + 16/3 + C
C = 2 - 1 - 16/3
C = -13/3
Maka,
y = x + 16/(3x³) - 13/3

Nomor 13.
a.)
a = t
v = ½ t² + v₀ = ½ t² + 2 -> v₂ = ½ (2)² + 2 = 4
s = ¹/₆ t³ + 2t + s₀
s = ¹/₆ t³ + 2t, karena s₀ = 0
s₂ = 1/6 (2)² + 2(2) = 14/3

b.)
a = (1+t)⁻³
v = -½ (1+t)⁻² + 4 -> v₂ = -½.(3)⁻² + 4 = 71/18
s = ½ (1+t)⁻¹ + 4t + 6
s₂ = ½ (3)⁻¹ + 4(2) + 6 = 83/6

c.)
a = (2t+1)^(1/3)
v = 3/8 (2t+1)^(4/3) -> v₂ = 3/8 ∛5⁴
s = 9/112 (2t+1)^(7/3) + 10
s₂ = 9/112 . ∛(5⁷) + 10

d.)
Sama dengan b, hanya saja.
s = 1/6 (1+t)⁻¹ + 4t
Menjadi:
v₂ tetap 71/18
s₂ = ½.(3)⁻¹ + 4(2) = 47/6