If the solution of the differential equation | vedclass

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(D) Given differential equation: $x \frac{d y}{d x} - y = x^2 \sin x$.Dividing by $x^2$,we get $\frac{x \frac{d y}{d x} - y}{x^2} = \sin x$.This is th

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$\alpha=\cot \frac{\alpha}{2}$

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(D) Given differential equation: $x \frac{d y}{d x} - y = x^2 \sin x$.Dividing by $x^2$,we get $\frac{x \frac{d y}{d x} - y}{x^2} = \sin x$.This is the derivative of $\frac{y}{x}$,so $\frac{d}{d x} \left( \frac{y}{x} ight) = \sin x$.Integrating both sides: $\frac{y}{x} = -\cos x + c$.Given $y(\pi) = 0$,we have $\frac{0}{\pi} = -\cos(\pi) + c \Rightarrow 0 = 1 + c \Rightarrow c = -1$.Thus,$y = -x \cos x - x$.For an extreme value at $x = \alpha$,we set $\frac{d y}{d x} = 0$.$\frac{d y}{d x} = -(\cos x - x \sin x) - 1 = -\cos x + x \sin x - 1 = 0$.At $x = \alpha$,$x \sin x - \cos x - 1 = 0$.Using trigonometric identities: $\alpha \sin \alpha - (\cos \alpha + 1) = 0$.$\alpha (2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}) - (2 \cos^2 \frac{\alpha}{2}) = 0$.$2 \cos \frac{\alpha}{2} (\alpha \sin \frac{\alpha}{2} - \cos \frac{\alpha}{2}) = 0$.Since $\cos \frac{\alpha}{2} 
eq 0$ for the existence of the function,we have $\alpha \sin \frac{\alpha}{2} = \cos \frac{\alpha}{2}$,which implies $\alpha = \cot \frac{\alpha}{2}$.

If the solution of the differential equation $x y^{\prime}=y+x^2 \sin x$ subject to the condition $y(\pi)=0$ is $y=f(x)$ and $f(x)$ has an extreme value at $x=\alpha$,then

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