Determine the longest interval in which the given initial value

Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the slotuion.
$$\displaystyle{\left({x}-{2}\right)}{y}{''}+{y}'+{\left({x}-{2}\right)}{\left({\tan{{x}}}\right)}{y}={0}$$
$$\displaystyle{y}{\left({3}\right)}={1}$$
$$\displaystyle{y}'{\left({3}\right)}={6}$$

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Marian Tucker
The given IVP
$$\displaystyle{\left({x}-{2}\right)}{y}{''}+{y}'+{\left({x}-{2}\right)}{\left({\tan{{x}}}\right)}{y}={0}$$
$$\displaystyle\Rightarrow{y}{''}+{\frac{{{1}}}{{{x}-{2}}}}{y}'+{\left({\tan{{x}}}\right)}{y}={0}$$
And the initial condition. $$\displaystyle{y}{\left({3}\right)}={1},\ {y}'{\left({3}\right)}={6}$$
Here $$\displaystyle{P}{\left({x}\right)}={\frac{{{1}}}{{{x}-{2}}}}$$
$$\displaystyle{q}{\left({x}\right)}={\tan{{x}}}$$
$$\displaystyle{g{{\left({x}\right)}}}={0}$$ constant function is continuous everywhere.
Solution: $$\displaystyle{\left({\left({2}{n}+{1}\right)}{\frac{{\pi}}{{{2}}}},{\left({2}{n}+{3}\right)}{\frac{{\pi}}{{{2}}}}\right)}$$