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…
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 $\cos y=$
$\frac{3}{\mathrm{x}^2}$
$\frac{1}{x}$
$\frac{1}{x^2}$
$\frac{2}{x^3}$
Solution
$\because \quad x^3 \sin y \frac{d y}{d x}=2 \Rightarrow \sin y d y=\frac{2}{x^3} d x$
Integrating both sides, we get
$\int \sin y d y=2 \int \frac{1}{x^3} d x \Rightarrow-\cos y=2 \cdot \frac{x^{-2}}{-2}+C$
$\Rightarrow \quad \cos y=\frac{1}{x^2}-C$ ...(i)
Taking $\lim _{x \rightarrow \infty}$ on both sides :
$\begin{aligned} & \Rightarrow \quad \lim _{x \rightarrow \infty} \cos y=\lim _{x \rightarrow \infty} \frac{1}{x^2}-C \\ & \Rightarrow \quad \cos \left(\lim _{x \rightarrow \infty} y\right)=0-C \Rightarrow \cos \left(\frac{\pi}{2}\right)=-C \Rightarrow C=0\end{aligned}$
Putting the value of $C$ in equation (i), we get :
$\Rightarrow \quad \cos y=\frac{1}{x^2}-0 \Rightarrow \cos y=\frac{1}{x^2}$