In the following questions, a statement of assertion (Statement-1) is followed by a statement of reason (Statement-2). Mark the correct choice as :Statement 1: xsin⁡xdydx+(x+xcos⁡x+sin⁡x)y=sin⁡x, yπ2=1−2π⇒limx→0 y(x)=13Statement 2: The differential equation is linear with integrating factor x(1 - cos x)

dydx+1sin⁡x+cot⁡x+1xy=1xI.F. =exp⁡∫1sin⁡x+cot⁡x+1xdx=exp ln⁡xtan⁡x2sin⁡x=xtan⁡x2×2sin⁡x2cos⁡x2=x(1−cos⁡x)Solution, yx(1−cos⁡x)=∫1x⋅x(1−cos⁡x)dx=x−sin⁡x+cyπ2=1−2π⇒c=0∴ y(x)=x−sin⁡xx(1−cos⁡x)=x−x−x36+…..x1−1−x22+…..=x261x22=13 as x→0.

In the following questions, a statement of assertion (Statement-1) is followed by a statement of reason (Statement-2). Mark the correct choice as :Statement 1: xsin⁡xdydx+(x+xcos⁡x+sin⁡x)y=sin⁡x, yπ2=1−2π⇒limx→0 y(x)=13Statement 2: The differential equation is linear with integrating factor x(1 - cos x)

dydx+1sin⁡x+cot⁡x+1xy=1xI.F. =exp⁡∫1sin⁡x+cot⁡x+1xdx=exp ln⁡xtan⁡x2sin⁡x=xtan⁡x2×2sin⁡x2cos⁡x2=x(1−cos⁡x)Solution, yx(1−cos⁡x)=∫1x⋅x(1−cos⁡x)dx=x−sin⁡x+cyπ2=1−2π⇒c=0∴ y(x)=x−sin⁡xx(1−cos⁡x)=x−x−x36+…..x1−1−x22+…..=x261x22=13 as x→0.

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In the following questions, a statement of assertion (Statement-1) is followed by a statement of reason (Statement-2). Mark the correct choice as :
Statement 1: $x \sin x \frac{d y}{d x} + (x + x \cos x + \sin x) y = \sin x$ , $y (\frac{π}{2}) = 1 - \frac{2}{π} \Rightarrow lim_{x \to 0} y (x) = \frac{1}{3}$
Statement 2: The differential equation is linear with integrating factor $x (1 - \cos x)$

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If both Statement-1 and Statement-I are true and Statement-2 is the correct explanation of Statement-1.

If both Statement-1 and Statement-2 are true but Statement-2 is not the correct explanation of Statement-1

If Statement-1 is true but Statement-2 is false

If Statement-1 is false but Statement-2 is true.

answer is A.

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Detailed Solution

$\frac{d y}{d x} + (\frac{1}{\sin x} + \cot x + \frac{1}{x}) y = \frac{1}{x}$

I.F. $= \exp \int (\frac{1}{\sin x} + \cot x + \frac{1}{x}) d x =$ exp $\ln (x \tan \frac{x}{2} \sin x)$

$= x \tan \frac{x}{2} \times 2 \sin \frac{x}{2} \cos \frac{x}{2} = x (1 - \cos x)$

Solution, $y x (1 - \cos x) = \int \frac{1}{x} \cdot x (1 - \cos x) d x = x - \sin x + c$

$\begin{array}{l} y (\frac{π}{2}) = 1 - \frac{2}{π} \Rightarrow c = 0 \\ ∴ y (x) = \frac{x - \sin x}{x (1 - \cos x)} \\ = \frac{x - (x - \frac{x^{3}}{6} + \dots . .)}{x (1 - (1 - \frac{x^{2}}{2} + \dots . .))} = \frac{x^{2}}{6} \frac{1}{\frac{x^{2}}{2}} = \frac{1}{3} as x \to 0 . \end{array}$