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Formation of a Differential Equation from a General Solution

If y cos x +x...

Question

If $y c o s x + x c o s y = π, t h e n y^{''} (0)$ is

A

1

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B

π

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C

\N

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D

- π

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Solution

The correct option is B $π$
$y c o s x + x c o s y = π$
Differentiate both sides with respect to x, we get
$- y s i n x + c o s x . y^{'} + x (- s i n y) y^{'} + c o s y$
Again differentiate with respect to x
$- y " s i n x - y c o s x + c o s x . y " + s i n x . y^{'} - s i n y . y^{'} - x [c o s y . (y^{'})^{2} + s i n y . y "] - s i n y . y^{'}$
Putting $x = 0, w e g e t - y + y " - 2 s i n y y^{'} = 0$
$y "= y + 2 y^{'} s i n y$
Since at $x = 0, y = π; (y ")_{0} = π$

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Similar questions

y cos x + x cos y = π,

y^{''} (0)

x, y, z

be real numbers with

x \geq y \geq z \geq \frac{π}{12}

x + y + z = \frac{π}{2}

p = cos x sin y cos z,

{cos}^{- 1} x + {cos}^{- 1} y = \frac{π}{2}

, then prove that

{cos}^{- 1} x = {sin}^{- 1} y

0 < x, y < π

cos x + cos y - cos (x + y) = \frac{3}{2},

sin x + cos y

y = y (x)

and it follows the relation

x cos y + y cos x = π

y^{^{''}} (0) = ?

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Formation of a Differential Equation from a General Solution

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