If lim _{x rightarrow infty} y(x)=frac{pi}{2} | vedclass

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(C) Given the differential equation: $x^3 \sin y \frac{d y}{d x}=2$. Separating the variables,we get: $\sin y \, dy = \frac{2}{x^3} \, dx$. Integratin

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$\frac{1}{x^2}$

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(C) Given the differential equation: $x^3 \sin y \frac{d y}{d x}=2$. Separating the variables,we get: $\sin y \, dy = \frac{2}{x^3} \, dx$. Integrating both sides: $\int \sin y \, dy = \int 2x^{-3} \, dx$. This yields: $-\cos y = 2 \cdot \frac{x^{-2}}{-2} + C = -\frac{1}{x^2} + C$. Multiplying by $-1$,we get: $\cos y = \frac{1}{x^2} - C$ ... $(i)$. Given the condition $\lim _{x \rightarrow \infty} y(x) = \frac{\pi}{2}$,we take the limit as $x \rightarrow \infty$ on both sides of equation $(i)$: $\lim _{x \rightarrow \infty} \cos y = \lim _{x \rightarrow \infty} \left( \frac{1}{x^2} - C \right)$. Since $\cos$ is a continuous function,$\cos(\lim _{x \rightarrow \infty} y) = 0 - C$. $\cos(\frac{\pi}{2}) = -C \Rightarrow 0 = -C \Rightarrow C = 0$. Substituting $C = 0$ into equation $(i)$,we get: $\cos y = \frac{1}{x^2}$.

If $\lim _{x \rightarrow \infty} y(x)=\frac{\pi}{2}$,then the solution of $x^3 \sin y \frac{d y}{d x}=2$ is $\cos y=$

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