I need help with this:
Given: [x - (sec y)(ln x)] dy + [tan y (y/x)sec y] dx = 0
Required: General Solution
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Suppose that M = tan y (y/x) sec y and N = x (sec y) (ln x)
Then, partial derivative of N wrt x = 1 ((sec y)/x)
And partial derivative of M wrt y = sec y (sec y ((y tan y)/x) - 1/x)
f(y) = - 1/M (sec y (sec y ((y tan y)/x) 1/x))
from there, to get the integrating factor,
f(y) = (sec^2 y ((y sec y tan y)/x) - 1) / (tan y (y/x)sec y)
and then, im kind of stuck.
oh yeah, it's positive. sorry.
back to the problem,
Suppose that M = tan y (y/x) sec y and N = x (sec y) (ln x)
Then, partial derivative of N wrt x = 1 ((sec y)/x)
And partial derivative of M wrt y = sec y (sec y ((y tan y)/x) - 1/x)
f(y) = - 1/M (sec y (sec y ((y tan y)/x) 1/x))
from there, to get the integrating factor,
f(y) = (sec^2 y ((y sec y tan y)/x) - 1) / (tan y (y/x)sec y)
and then, im kind of stuck.