The solution of $x c o s^{2} y (d x) + t a n y (d y) = 0$ is:

x^{2} + s e c^{2} y = c

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x^{2} + c o t^{2} y = c

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x^{2} + s i n^{2} y = c

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x^{2} + c o s^{2} y = c

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Solution

The correct option is A $x^{2} + s e c^{2} y = c$
$x c o s^{2} y d x + t a n y d y = 0$
$\Rightarrow x d x + t a n y s e c^{2} y d y = 0$
Let tan y = t
$\Rightarrow s e c^{2} y d y = d t$
$\int x d x + \int t d t = c_{1}$
$\Rightarrow \frac{x^{2}}{2} + \frac{t^{2}}{2} = 2 c_{1}$
$x^{2} + t a n^{2} y = 2 c_{1}$
$x^{2} + s e c^{2} y = 2 c_{1} + 1$ put $(2 c_{1} + 1) = c$
$\Rightarrow x^{2} + s e c^{2} y = c$

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Similar questions

The solution of differential equation $\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{(1 + x^{2})^{2}}$ is
(a) $y (1 + x^{2}) = C + t a n^{- 1} x$
(b) $\frac{y}{1 + x^{2}} = C + t a n^{- 1} x$
(c) $y l o g (1 + x^{2}) = C + t a n^{- 1} x$
(d) $y (1 + x^{2}) = C + s i n^{- 1} x$