Proving a differential equation using the substitution method

[jeffreylze]

Homework Statement

dy/dx = \frac{[2cos(x)^2-sin(x)^2+y^2]}{[2cos(x)]}

Substitute y(x) = sin(x) + \frac{1}{u(x)}

Homework Equations

By doing the substitution, it will yield the differential equation for u(x)

du/dx = -u tan(x) - \frac{1}{2}sec(x)

The Attempt at a Solution

I figured out i have to use chain rule. However, if du/dx = du/dy x dy/dx , which dy/dx do i choose? It can be either

this - dy/dx = \frac{[2cos(x)^2-sin(x)^2+y^2]}{[2cos(x)]}

or - y = sin(x) + 1/[u(x)]
dy/dx = cos(x)

Then, I found the dy/du from this equation y = sin(x) + 1/[u(x)] and flipped it around to get du/dy. After multiplying using the chain rule, I don't get the differential equation as shown. Please help me out here. =X

 

[jeffreylze]

I just realized a similar question has been posted. Please close the thread. Sorry for the trouble. =X