Jawabanmu

2014-06-06T07:46:25+07:00
(2y - 1) dy/dx = 3x² + 1
(2y - 1) dy = (3x² + 1) dx
Integral (2y - 1) dy = Integral (3x² + 1) dx
y² - y + c1 = x³ + x + c2 <----- kedua ruas dikalikan 4
4y² - 4y + 4c1 = 4x³ + 4x + 4c2
(2y - 1)² + 4c1 = 4x³ + 4x + 1 + 4c2
(2y - 1)²= 4x³ + 4x + 1 + c
2y - 1 = ± √(4x³ + 4x + 1 + c)
2y = 1 ± √(4x³ + 4x + 1 + c)
y = 1/2 (1 ± √(4x³ + 4x + 1 + c))
2014-06-06T07:57:25+07:00
y = 1/2 (1 ± √(4x³ + 4x + 1 + c))