# 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 solution.
$$\displaystyle{\left({x}+{3}\right)}{y}{''}+{x}{y}'+{\left({\ln}{\left|{x}\right|}\right)}{y}={0},\ {y}{\left({1}\right)}={0},\ {y}'{\left({1}\right)}={1}$$

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Mike Henson
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 $$\displaystyle{\left({0},\infty\right)}$$