TUT 0202 QUIZ1

[Yijin Qiang]

${y}'+y^{2}sin(x)= 0\\\frac{dy}{dx}= -y^{2}sin(x)\\dy=-y^{2}sin(x)dx\\-\int \frac{1}{y^{2}}dy=\int sin(x)dx\\-\frac{1}{y}=-cos(x)+C\\y=\frac{1}{cos(x)+C}$