We have the differential equation

\(\displaystyle{\left({x}+{3}\right)}{y}{''}+{x}{y}'+{\left({\ln}{\left|{x}\right|}\right)}{y}={0}\)

which is equivalent to

\(\displaystyle{y}{''}+{\frac{{{x}}}{{{x}+{3}}}}{y}'+{\frac{{{\ln}{\left|{x}\right|}}}{{{x}+{3}}}}{y}={0}\)

as the form

\(\displaystyle{y}{''}+{p}{\left({x}\right)}{y}'+{q}{\left({x}\right)}{y}={g{{\left({x}\right)}}}\)

which the initial values

\(\displaystyle{y}{\left({1}\right)}={0},{y}'{\left({1}\right)}={1}\)

Using the notation of Theorem, we find that

1, for \(\displaystyle{p}{\left({x}\right)}={\frac{{{x}}}{{{x}+{3}}}}\), it is continuous on \(\displaystyle{\left(-\infty,-{3}\right)}\) and \(\displaystyle{\left(-{3},\infty\right)}\)

2. for \(\displaystyle{q}{\left({x}\right)}={\frac{{{\ln}{\left|{x}\right|}}}{{{x}+{3}}}}\), it is continuous on \(\displaystyle{\left(-\infty,-{3}\right)},{\left(-{3},{0}\right)}\) and \(\displaystyle{\left({0},\infty\right)}\).

3. for \(\displaystyle{g{{\left({x}\right)}}}={0}\), it is continuous on \(\displaystyle{\left(-\infty,\infty\right)}\)

Since we have \(\displaystyle{t}_{{0}}={1}\), then the longest interval of the solution is \(\displaystyle{\left({0},\infty\right)}\)

\(\displaystyle{\left({x}+{3}\right)}{y}{''}+{x}{y}'+{\left({\ln}{\left|{x}\right|}\right)}{y}={0}\)

which is equivalent to

\(\displaystyle{y}{''}+{\frac{{{x}}}{{{x}+{3}}}}{y}'+{\frac{{{\ln}{\left|{x}\right|}}}{{{x}+{3}}}}{y}={0}\)

as the form

\(\displaystyle{y}{''}+{p}{\left({x}\right)}{y}'+{q}{\left({x}\right)}{y}={g{{\left({x}\right)}}}\)

which the initial values

\(\displaystyle{y}{\left({1}\right)}={0},{y}'{\left({1}\right)}={1}\)

Using the notation of Theorem, we find that

1, for \(\displaystyle{p}{\left({x}\right)}={\frac{{{x}}}{{{x}+{3}}}}\), it is continuous on \(\displaystyle{\left(-\infty,-{3}\right)}\) and \(\displaystyle{\left(-{3},\infty\right)}\)

2. for \(\displaystyle{q}{\left({x}\right)}={\frac{{{\ln}{\left|{x}\right|}}}{{{x}+{3}}}}\), it is continuous on \(\displaystyle{\left(-\infty,-{3}\right)},{\left(-{3},{0}\right)}\) and \(\displaystyle{\left({0},\infty\right)}\).

3. for \(\displaystyle{g{{\left({x}\right)}}}={0}\), it is continuous on \(\displaystyle{\left(-\infty,\infty\right)}\)

Since we have \(\displaystyle{t}_{{0}}={1}\), then the longest interval of the solution is \(\displaystyle{\left({0},\infty\right)}\)