The interval contained in the domain of definition of non-zero solutions of the differential equation $(x-3)^2 y'+y=0$, is
Answer & explanation
Correct answer: option 1
We have,
$(x-3)^2 y'+y=0$
$\Rightarrow (x-3)^2 \frac{d y}{d x}+y=0 \Rightarrow \frac{1}{y} d y+\frac{1}{(x-3)^2} d x=0$
Integrating, we get
$\log |y|-\frac{1}{x-3}=\log C$
$\Rightarrow \log \left\{\frac{|y|}{C}\right\}=\frac{1}{x-3} \Rightarrow|y|=C e^{\frac{1}{x-3}}$
The domain of definition of this solution is $R-\{3\}$.
Clearly, $(-\pi / 2, \pi / 2) \subset R-\{3\}$. So, option (a) is correct.