An engineer is modeling the cross-sectional profile of a specialized optical lens. The thickness of the lens, y y\,y mm, at a horizontal distance x x\,x mm from the optical axis, satisfies a specific differential equation.
Find the derivative with respect to y y\,y of
1(1+2lny)2 \frac{1}{(1 + 2\ln y)^2} (1+2lny)21Hence find a general solution to the differential equation
12csc(2x)dydx=y(1+2lny)3 12\csc(2x) \frac{dy}{dx} = y(1 + 2\ln y)^3 12csc(2x)dxdy=y(1+2lny)3for y>0 y > 0\,y>0 and −π2<x<π2\displaystyle -\frac{\pi}{2} < x < \frac{\pi}{2}−2π<x<2π.
Show that the particular solution of this differential equation for which y=e1/2y = e^{1/2}y=e1/2 at x=π6\displaystyle x = \frac{\pi}{6}x=6π is given by
y=eAsecx−12 y = e^{A\sec x - \frac{1}{2}} y=eAsecx−21where A A\,A is a constant to be found.
