Let y = y(x) be the solution curve of the differential equation sec y d y d x + 2 x sin y = x 3 cos y , y(1) = 0. Then y ( 3 ) is equal to: [2024]

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Solution

Q.
Let y = y(x) be the solution curve of the differential equation $\sec y \frac{d y}{d x} + 2 x \sin y = x^{3} \cos y$ , y(1) = 0. Then $y (\sqrt{3})$ is equal to: [2024]

1 $\frac{π}{6}$

2 $\frac{π}{3}$

3 $\frac{π}{4}$

4 $\frac{π}{12}$

Ans.
(3)

We have, $\sec y \frac{d y}{d x} + 2 x \sin y = x^{3} \cos y$

$\Rightarrow \sec^{2} y \frac{d y}{d x} + 2 x \tan y = x^{3}$ ... (i)

Let $z = \tan y \Rightarrow \frac{d z}{d x} = \sec^{2} y \frac{d y}{d x}$

$∴ From (i), \frac{d z}{d x} + 2 x z = x^{3}$

$IF = e^{\int 2 x d x} = e^{x^{2}} \Rightarrow z \cdot e^{x^{2}} = \int e^{x^{2}} \cdot x^{3} d x + c$

$\Rightarrow \tan y \cdot e^{x^{2}} = \frac{1}{2} (x^{2} e^{x^{2}} - e^{x^{2}}) + c$

Since, y(1) = 0

$∴ \tan (0) \cdot e = \frac{1}{2} (1 \cdot e - e) + c \Rightarrow c = 0$

So, $\tan y e^{x^{2}} = \frac{1}{2} (x^{2} e^{x^{2}} - e^{x^{2}})$

$\Rightarrow \tan y = \frac{x^{2} - 1}{2} \Rightarrow y = \tan^{- 1} (\frac{x^{2} - 1}{2})$

$∴ y (\sqrt{3}) = \tan^{- 1} (\frac{{(\sqrt{3})}^{2} - 1}{2}) = \tan^{- 1} (1) = \frac{π}{4}$

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