A curve passes through the point \left(\right. 1 , \textrm{ } \frac{\pi}{6} \left.\right) . Let the slope of the curve at each point (x, y) be \frac{y}{x} + sec \textrm{ } \left(\right. \frac{y}{x} \left.\right) , x

Engineering

Mathematics

Homogeneous Differential Equation

Question

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

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

$\sec (\frac{2 y}{x}) = logx + 2$

$cosec (\frac{y}{x}) = logx + 2$

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

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$Given, \frac{dy}{dx} = \frac{y}{x} + \sec (\frac{y}{x})$ ...........(1)

Put y = mx

$∴ \frac{dy}{dx} = m + x \frac{dm}{dx}$ .............(2)

From (1) and (2), we get $\frac{dm}{secm} = \frac{dx}{x}$

$∴ \int cosm dx = ℓnx + λ \Rightarrow sinm = ℓnx + λ$

$Put (1, \frac{π}{2}), we get λ = \frac{1}{2}$

$∴ \sin (\frac{y}{x}) = ℓnx + \frac{1}{2}$