If a curve y = y(x) passes through the point and satisfies the differential equation , then at x = 2, the value of cos y is: [2025]

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Solution

Q.
If a curve y = y(x) passes through the point $(1, \frac{π}{2})$ and satisfies the differential equation $(7 x^{4} \cot y - e^{x} cosec y) \frac{d x}{d y} = x^{5}, x \geq 1$ , then at x = 2, the value of cos y is: [2025]

1 $\frac{2 e^{2} + e}{128}$

2 $\frac{2 e^{2} - e}{128}$

3 $\frac{2 e^{2} - e}{64}$

4 $\frac{2 e^{2} + e}{64}$

Ans.
(2)

We have, $(7 x^{4} \cot y - e^{x} cosec y) \frac{d x}{d y} = x^{5}$

$\Rightarrow \frac{d y}{d x} = \frac{7 x^{4} \cot y}{x^{5}} - \frac{e^{x} cosec y}{x^{5}}$

$\Rightarrow \frac{d y}{d x} - \frac{7}{x} \cot y = \frac{- e^{x}}{x^{5}} cosec y$

$\Rightarrow \sin y \cdot \frac{d y}{d x} - \cos y (\frac{7}{x}) = - \frac{e^{x}}{x^{5}}$

If $- \cos y = t \Rightarrow \sin y \cdot \frac{d y}{d x} = \frac{d t}{d x}$

$\Rightarrow \frac{d t}{d x} + t (\frac{7}{x}) = - \frac{e^{x}}{x^{5}}$

Here, $I . F . = x^{7}$

$\Rightarrow t \cdot x^{7} = - \int x^{2} \cdot e^{x} d x \Rightarrow \cos y \cdot x^{7} = x^{2} e^{x} - 2 \int x e^{x} d x$

$\Rightarrow \cos y \cdot x^{7} = x^{2} e^{x} - 2 x e^{x} + 2 e^{x} + c$

at point $(1, \frac{π}{2})$ , we get c = –e

Thus, the value of cos y at x = 2 is

$\cos y = \frac{2 e^{2} - e}{128}$

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