Interval contained in the domain of definition of non-zero solution of the differential equation `(x-3)^(2)y'+y=0` is:

Options

A. `(-pi//2,pi//2)`
B. `(0,pi)`
C. `(0,2pi)`
D. `(-pi,pi)`

Answer

A

Explanation

We have,
`(x-3)^(2)y'+y=0`
`rArr" "(x-3)^(2)(dy)/(dx)+y=0rArr(1)/(y)dy+(1)/((x-3)^(2))dx=0`
Integrating, we get
`log|y|-(1)/(x-3)=logC`
`rArr" "log{(|y|)/(C)}=(1)/(x-3)rArr|y|=Ce^((1)/(x-3))`
The domain of definition of this solution is `R-{3}`.
Clearly, `(-pi//2,pi//2)subR-{3}`. So, option (a) is correct .