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

Explanation

According to the given condition slope of tangent, `m=frac{dy}{dx}`

`frac{dy}{dx}=y/x-cos^2 frac{y}{x}`

Let y=vx,

`v+x frac{dv}{dx}=v-cos^2v=>x farc{dv}{dx}=-cos^2v`

`sec^2v dv=-frac{dx}{v}=>tan v=-logx+c int sec^2 v dv=int frac{-dx}{x}`

`=tan y/x+log x=C`

`x=1, y= pi/4` we get `C=1`

`tan y/x+log x=1`