Determine the longest interval in which the given initial value

smismSitlougsyy 2021-11-11 Answered
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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Expert Answer

Mike Henson
Answered 2021-11-12 Author has 8249 answers
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 \(\displaystyle{\left({0},\infty\right)}\)
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