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[y² + 3y] dy = [x² + 3x⁻¹] dx
integralkan kedua ruas
∫ [y² + 3y] dy = ∫ [x² + 3x⁻¹] dx
1/3.y³ + 3/2.y² + C₁ = 1/3.x³ 3㏑|x| + C₂
1/3.y³ + 3/2.y² + C₁ = x³.㏑|x| + C₂
1/3.y³ + 3/2.y² - x³.㏑|x| + C₂ - C₁ = 0
∴ solusi umum 1/3.y³ + 3/2.y² - x³.㏑|x| + C = 0