Answer to Question #351041 in Differential Equations for Taras

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

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