Let y=y(x) solve dxdy=sinx(secx−sinxtanx)tanx+y, x∈(0,2π), y(4π)=2. Then y(3π) is:
Solutions · 1
Official solution
Simplifying the RHS and solving the resulting linear ODE (dividing through appropriately) gives y=xsecx(…); applying y(π/4)=2 fixes the constant, and evaluating at π/3 gives y=3(2+loge3).