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: \(\displaystyle{\left({\left({2}{n}+{1}\right)}{\frac{{\pi}}{{{2}}}},{\left({2}{n}+{3}\right)}{\frac{{\pi}}{{{2}}}}\right)}\)

\(\displaystyle{\left({x}-{2}\right)}{y}{''}+{y}'+{\left({x}-{2}\right)}{\left({\tan{{x}}}\right)}{y}={0}\)

\(\displaystyle\Rightarrow{y}{''}+{\frac{{{1}}}{{{x}-{2}}}}{y}'+{\left({\tan{{x}}}\right)}{y}={0}\)

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: \(\displaystyle{\left({\left({2}{n}+{1}\right)}{\frac{{\pi}}{{{2}}}},{\left({2}{n}+{3}\right)}{\frac{{\pi}}{{{2}}}}\right)}\)