If $\frac{d y}{d x} + 2 x tan (x - y) = 1 t h e n f i n d t h e v a l u e o f s i n (x - y)$ ___.
(Here A is a constant)

A e^{- x^{2}}

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A e^{2 x}

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A e^{x^{2}}

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A e^{- 2 x}

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Solution

The correct option is D $A e^{x^{2}}$
Take $x - y = t$

$\Rightarrow 1 - \frac{d y}{d x} = \frac{d t}{d x}$

$\frac{d y}{d x} + 2 x tan (x - y) = 1$
$\Rightarrow 2 x tan t = \frac{d t}{d x}$
$\Rightarrow 2 x d x = \frac{cos t d t}{sin t}$

$x^{2} = log (sin t) + C$

$\Rightarrow sin t = A e^{x^{2}}$

$\Rightarrow sin (x - y) = A e^{x^{2}}$

Suggest Corrections

Similar questions

y = a e^{- 1 / x} + b

\frac{d y}{d x} = \frac{y}{x^{2}}

y = a e^{- 1 / x} + b

\frac{d y}{d x} = \frac{y}{x^{2}}

y^{2} = a e^{- 2 x} + (2 / 5) (c o s x - 2 s i n x)

y \frac{d y}{d x} + y^{2} + s i n x

y = a e^{2 x} + b e^{- x}

\frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 2 y = 0

y = 2 sec x

\frac{d y}{d x}

2 sec x tan x

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