Question

2.2. Solve the ivp sin(x) dx + y dy = 0, where y(0)

Answer 1

$$\sin(x) dx + y dy = 0, \quad y(0)=-2.$$

$$y dy = -\sin(x) dx$$

$$y^2/2=\cos(x)+C$$

$$2=\cos(0)+C=1+C$$

$$C=1$$

Finally

$$y^2=2\cos(x)+2$$

$$y=\pm\sqrt{2\cos(x)+2}$$

The initial condition requres negative value of $y$ for specified $x$, thus

$$y=-\sqrt{2\cos(x)+2}$$