Trivial and non-trivial solutions of second order differential.

Question

Show that $y=x^2\sin(x)$ and $y=0$ are both solutions of $$x^2y''-4xy'+(x^2+6)y=0$$ and that both satisfy the conditions $y(0) = 0$ and $y'(0) = 0$. Does this theorem contradict Theorem A? If not, why not?

Theorem A states that there exists only one function satisfying the homogeneous second order differential for a particular set of values of $y(x_•),y'(x_•)$ .

What I did:-

1) checked validity of solutions.

2) found another independent solution apart from $x^2\sin(x)$ i.e. $x^2\cos(x)$ and hence the general solution of the homogeneous second order linear differential.

What I want know:

1) if the theorem demands the requirement of non-trivial sol'n...or how exactly can I interpret it (preferably in simple words since at my level the proof of the theorem is not disclosed)

2) or if the above case actually goes against the theorem (unlikely, I guess...)

3) Any good sources to understand all these nuances, since this question is making it difficult for me to understand another theory on normal form of second order homogenous linear equation.

Thanks in advance!

Answer 1

The existence and uniqueness is usually stated when the coefficient of $y''$ is $1$, so you want to look at $$y''(x) - \frac{4}{x}y'(x) + \left(1+\frac{6}{x^2}\right)y(x) = 0.$$There is no contradiction because the coefficients $$-\frac{4}{x}\quad\mbox{and}\quad 1+\frac{6}{x^2}$$are continuous except at $x=0$.

Answer 2

Making

$$ y = x^{\alpha}e^{\pm i x} $$

and substituting we obtain

$$ (\alpha-2) e^{i x} (\alpha\pm2 i x-3) x^{\alpha} = 0 $$

hence with $\alpha = 2$ and

$$ y = x^2(C_1\sin x+ C_2\cos x) $$

is the general solution