Question
If $\vec{\text{a}}$ be the position vector whose tip is (5, -3), find the coordinates of a point B such that $\overrightarrow{\text{AB}}=\vec{\text{a}}$, the coordinates of A being (4, -1).
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$\text{x}\sqrt{1-\text{y}^2}\text{dx}+\text{y}\sqrt{1-\text{x}^2}\text{dy}=0$
$\text{f(x)}=\begin{cases}(\text{x}-\text{c})\cos\Big(\frac{1}{\text{x}-\text{c}}\Big), & \text{x}\neq 0\\0, & \text{x}= 0\end{cases}$
$x^{2} \frac{d^{2}y}{dx^{2}} + x \frac{dy}{dx} + y =0$