Trivial and non-trivial solutions of second order differential.

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.

• I don't read vertically, so I have no idea what your huge image says. Also, text in images cannot be found by future searchers, so you should type the image contents. You may be benefit from block quotation markup (start the line with a ">" ). Jun 20, 2018 at 18:31
• Sorry for the picture. I have edited it. Jun 20, 2018 at 18:35

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 $x=0$.

• Thanks a lot. I rechecked the statement of the theorem and with the help of your answer realised that there is a discontinuity at x=0, however, what about these two solutions are they both accurate ( allowed due to this discontinuity)? Jun 20, 2018 at 18:50
• Yes, both solutions are valid, because the initial conditions are given at $x=0$. If you had a initial condition at, say, $x=1$, you'd get a unique solution in $]0,+\infty[$. Jun 20, 2018 at 18:53

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

• I did say I calculated the above though...should I have been more clear? Jun 20, 2018 at 18:51