determination of linearly independent solutions

Question

If y_1(x)=x^2 and y_2(x)=(x^2 )logx then in what interval can both these functions be linearly independent. According to me it must be the whole real no. except 0. Am i right?

Answer 1

To be defined, as Adam noted, we need $x>0$, so I am presuming that.

Then the two functions are linearly independent on any set $S$ that contains at least two points.

If $a x^2 + b x^2 \log x = 0$ then since $x>0$ we have $a + b \log x = 0 $.

If $b=0$, then we must have $a=0$, so assume that $b \neq 0$. Then $ x = e^{- {a \over b}} $.

It is clear that this can hold for at most one value of $x$, hence if $S$ contains at least two points then $a=b= 0$.

In particular, on any interval (contained in $(0,\infty)$) the functions are linearly independent.

Answer 2

Let $I\subseteq(0,\infty)$ be a non-degenerate interval. Then $$ \alpha_1y_1(x)+\alpha_2y_2(x)=0\quad\forall x\in I $$ if and only if $$ \alpha_1x^2+\alpha_2x^2\log x=0\quad\forall x\in I $$ if and only if $$ \alpha_1+\alpha_2\log x=0\quad\forall x\in I $$ if and only if $$ \alpha_2\log x=-\alpha_1\quad\forall x\in I\tag{1} $$ Now, $-\alpha_1$ is a constant. So if $\alpha_2\neq0$ then $(1)$ can't be true for all $x\in I$ (taking two different values of $x$ will give two different values of $\log x$ by injectivity and hence different values of $\alpha_2\log x$ and $\alpha_2\log x$ won't be constant). Hence we conclude that $\alpha_2=0$ and it follows that $\alpha_1=0$ too. Hence $$ \alpha_1y_1(x)+\alpha_2y_2(x)=0\quad\forall x\in I\iff\alpha_1=\alpha_2=0 $$ and $y_1$ and $y_2$ are linearly independent on $I$.