The equation of the curve passing through the point `(1,pi/4)` and having a slope of tangent at any point (x,y) as `y/x - cos^2(y/x)` is

Options

A. ` x = e^(1+tan(y/x))`
B. ` x = e^(1-tan(y/x))`
C. ` x = e^(1+tan(x/y))`
D. ` x = e^(1-tan(x/y))`

Answer

c

Explanation

Given, `(dy)/(dx) = y/x - (cos^(2)) (y/x)`
` rArr (x-dy-y dx)/x = - (-cos^(2). y/x)dx`
` sec^(2) (y/x) ((x-dy-ydx)/(x^(2))) = -(dx)/2`
`rArr sec^(2). y/2 .d(y/x) = -(dx)/x`
On integrating both sides we get ltbRgt ` rArr tan. y/x = - log x+C`
when x = 1, `y = pi/4 rArr C = 1`
` :. tan.(y/x) = 1- log x rArr x = e^(1-tan.(y/x))`