Show that y=x^2\sin(x) and y=0 are both solutions of x^2y''-4xy'+(x^2+6)y=0 and that

Yolanda Jorge 2021-11-20 Answered
Show that \(\displaystyle{y}={x}^{{2}}{\sin{{\left({x}\right)}}}\) and \(\displaystyle{y}={0}\) are both solutions of
\(\displaystyle{x}^{{2}}{y}{''}-{4}{x}{y}'+{\left({x}^{{2}}+{6}\right)}{y}={0}\)
and that both satisfy the conditions \(\displaystyle{y}{\left({0}\right)}={0}\) and \(\displaystyle{y}'{\left({0}\right)}={0}\). Does this theorem contradict Theorem A? If not, why not?

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Expert Answer

Harr1957
Answered 2021-11-21 Author has 760 answers
The existence and uniqueness is usually stated when the coefficient of \(\displaystyle{y}{''}\) is 1, so you want to look at
\(\displaystyle{y}{''}{\left({x}\right)}-{\frac{{{4}}}{{{x}}}}{y}'{\left({x}\right)}+{\left({1}+{\frac{{{6}}}{{{x}^{{2}}}}}\right)}{y}{\left({x}\right)}={0}\)
There is no contradiction because the coefficients
\(\displaystyle-{\frac{{{4}}}{{{x}}}}\) and \(\displaystyle{1}+{\frac{{{6}}}{{{x}^{{2}}}}}\)
are continuous except at x=0
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Wasither1957
Answered 2021-11-22 Author has 1348 answers
Making
\(\displaystyle{y}={x}^{\alpha}{e}^{{\pm{i}{x}}}\)
and substituting we obtain
\(\displaystyle{\left(\alpha-{2}\right)}{e}^{{{i}{x}}}{\left(\alpha\pm{2}{i}{x}-{3}\right)}{x}^{\alpha}={0}\)
hence with \(\displaystyle\alpha={2}\) and
\(\displaystyle{y}={x}^{{2}}{\left({C}_{{1}}{\sin{{x}}}+{C}_{{2}}{\cos{{x}}}\right)}\)
is the general solution
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