A curve passes through the point (1,π/6) . Let the slope of the curve at each point (x,y) be yx+sec⁡(y/x),x>0 then the equation of the curve is

dydx=yx+sec⁡(y/x) put y=vxv+xdvdx=v+secv→dv secv =dxx→∫cosvdv=∫dxx→sin⁡(y/x)=ln⁡x+c The curve passes through (1,π/6) is →12=c→sin⁡(y/x)=ln⁡x+12

A curve passes through the point (1,π/6) . Let the slope of the curve at each point (x,y) be yx+sec⁡(y/x),x>0 then the equation of the curve is

dydx=yx+sec⁡(y/x) put y=vxv+xdvdx=v+secv→dv secv =dxx→∫cosvdv=∫dxx→sin⁡(y/x)=ln⁡x+c The curve passes through (1,π/6) is →12=c→sin⁡(y/x)=ln⁡x+12

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$A curve passes through the point (1, π / 6) . Let the slope of the curve at each$ $point (x, y) be \frac{y}{x} + \sec (y / x), x > 0 then the equation of the curve is$

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$\sin (y/x) = logx + \frac{1}{2}$

$cos e c (y/x) = logx + 2$

$s e c (2y/x) = logx + 2$

$cos (2y/x) = logx + \frac{1}{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{dy}{dx} = \frac{y}{x} + \sec (y / x) put y = v x \\ v + x \frac{d v}{d x} = v + s e c v \\ \to \frac{dv}{secv} = \frac{dx}{x} \to \int cosvdv = \int \frac{dx}{x} \\ \to \sin (y / x) = \ln x + c \\ The curve passes through (1, π / 6) is \end{array}$