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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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The correct answer is:
$\alpha=\cot \frac{\alpha}{2}$
Given differential equation,
$\begin{aligned} x \frac{d y}{d x}-y & =x^2 \sin x \\ \Rightarrow \quad \frac{x d y-y d x}{x^2} & =\sin x d x \Rightarrow \int d\left(\frac{y}{x}\right)=\int \sin x d x \\ \Rightarrow \quad \frac{y}{x} & =-\cos x+c\end{aligned}$
$\therefore \quad y(\pi)=0 \Rightarrow c=-1 \Rightarrow y=-x \cos x-x$
Now, $\frac{d y}{d x}=-\cos x+x \sin x-1=0$ for extremum.
$\Rightarrow \quad 2 x \sin \frac{x}{2} \cos \frac{x}{2}=2 \cos ^2 \frac{x}{2} \Rightarrow x=\cot \frac{x}{2}$
$\therefore$ Displacement of object, $\begin{aligned} \Delta x & =x_2-x_1 \\ & =(\alpha+16 \beta)-(\alpha+4 \beta) \\ & =12 \beta\end{aligned}$
Average velocity,
$\begin{array}{ll} & v_{\text {avg }}=\frac{\Delta x}{\Delta t}=\frac{12 \beta}{(4-2)} \\ \Rightarrow & 12=\frac{12 \beta}{2} \\ \Rightarrow & \beta=2 \mathrm{~ms}^{-2}\end{array} \quad\left(\because v_{\text {avg }}=12 \mathrm{~ms}^{-1}\right)$
$\begin{aligned} x \frac{d y}{d x}-y & =x^2 \sin x \\ \Rightarrow \quad \frac{x d y-y d x}{x^2} & =\sin x d x \Rightarrow \int d\left(\frac{y}{x}\right)=\int \sin x d x \\ \Rightarrow \quad \frac{y}{x} & =-\cos x+c\end{aligned}$
$\therefore \quad y(\pi)=0 \Rightarrow c=-1 \Rightarrow y=-x \cos x-x$
Now, $\frac{d y}{d x}=-\cos x+x \sin x-1=0$ for extremum.
$\Rightarrow \quad 2 x \sin \frac{x}{2} \cos \frac{x}{2}=2 \cos ^2 \frac{x}{2} \Rightarrow x=\cot \frac{x}{2}$
$\therefore$ Displacement of object, $\begin{aligned} \Delta x & =x_2-x_1 \\ & =(\alpha+16 \beta)-(\alpha+4 \beta) \\ & =12 \beta\end{aligned}$
Average velocity,
$\begin{array}{ll} & v_{\text {avg }}=\frac{\Delta x}{\Delta t}=\frac{12 \beta}{(4-2)} \\ \Rightarrow & 12=\frac{12 \beta}{2} \\ \Rightarrow & \beta=2 \mathrm{~ms}^{-2}\end{array} \quad\left(\because v_{\text {avg }}=12 \mathrm{~ms}^{-1}\right)$
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