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

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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Harr1957
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
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