49.The solutions of p+cos(px)sin(y)= sin(px)cos(y) ( where p= dy/dx) Options a)y=0 b) cx-y=arcsin(x) c)cx-y= arcsin(c) d)y=(x-1 )-arcsin((x-1)/x)

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p + cos p x sin y = sin p x cos y P = (\frac{d y}{d x})

y = x [\frac{d y}{d x}] + {[\frac{d y}{d x}]}^{3}

y = c x + \frac{1}{x}

y \frac{d y}{d x} = x {(\frac{d y}{d x})}^{2} + 1

(x + l o g y) d y + y d x = 0

y (0) = 1

\frac{d y}{d x} = \frac{1 - cos x}{1 + cos x}

\frac{d y}{d x} = \sqrt{4 - y^{2}} (- 2 < y < 2)

\frac{d y}{d x} = 1 (y \neq 1)

{sec}^{2} x tan y d x + {sec}^{2} y tan x d y = 0

(e^{x} + e^{- x}) d y - (e^{x} - e^{- x}) d x = 0

\frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})

y log y d x - x d y = 0

x^{4} \frac{d y}{d x} = - y^{5}

\frac{d y}{d x} = {sin}^{- 1} x

e^{x} tan y d x + (1 - e^{x}) {sec}^{2} y d y = 0

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