Question

# Solve differential equationsin(x) dy/dx+(cos(x))y=0, y((7pi)/6)=-2

First order differential equations

Solve differential equation $$\sin(x) dy/dx+(\cos(x))y=0$$, $$y((7\pi)/6)=-2$$

2020-11-15

$$\sin x dy/dx+ y\cos x=0$$
$$\sin x dy= -y\cos x dx$$
$$dy/y= - \cos x/\sin x dx$$
$$\int dy/y= -\int \cos x/\sin x dx$$
$$\ln |(y)|= -\ln |(\sin x)|+\ln |(C)|$$
$$\ln y= \ln (C/\sin x)$$
$$y\sin x= C$$
Now, We are applying the given Initial Condition is as follow
$$y((7\pi)/6)= -2$$
$$-2*\sin((7\pi)/6)= C$$ $$-2* -1= C$$ $$\therefore\sin(7\pi/6)= -1$$
C=2
$$y\sin x= 2$$