- #1
jeffreylze
- 44
- 0
Homework Statement
dy/dx = [tex]\frac{[2cos(x)^2-sin(x)^2+y^2]}{[2cos(x)]}[/tex]
Substitute y(x) = sin(x) + [tex]\frac{1}{u(x)}[/tex]
Homework Equations
By doing the substitution, it will yield the differential equation for u(x)
du/dx = -u tan(x) - [tex]\frac{1}{2}[/tex]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 = [tex]\frac{[2cos(x)^2-sin(x)^2+y^2]}{[2cos(x)]}[/tex]
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