Let $y_1(x)=x^3$ and $y_2(x)=x^2|x|$ for $x\in \mathbb{R}$. Consider the following statements:
$(P)$ $y_1(x)$ and $y_2(x)$ are linearly independent solutions of $x^2\frac{d^2y}{dx^2}-4x\frac{dy}{dx}+6y=0$ on $\mathbb{R}$.
$(Q)$ The Wronskian $y_1(x)\frac{dy_2}{dx}(x)-y_2(x)\frac{dy_1}{dx}(x)=0$ for all $x\in \mathbb{R}$.
Then which of the above statements hold true.
The question appears in GATE-$2017$. I am trying this problem in following steps:
Both $y_1(x)$ and $y_2(x)$ are solutions: First I checked that $y_1(x)$ is a solution of above ODE. For $y_2(x)$, I checked that it is differentiable and $y_2(x)=x^3$ is a solution for $x\geq0$ and $y_2(x)=-x^3$ is a solution for $x<0$ of the above ODE.
Find their Wronskian: For $x\geq0$; $w(y_1,y_2)=w(x^3,x^3)=0$ and for $x<0$; $w(y_1,y_2)=w(x^3,-x^3)=0$. Thus Wronskian of $y_1(x)$ and $y_2(x)$ is zero for all $x\in \mathbb{R}$.
Since $y_1(x)$ and $y_2(x)$ both are solutions of the above ODE and their Wronskian is zero for all $x\in \mathbb{R}$ then can we conclude $y_1(x)$ and $y_2(x)$ are linealy dependent? But the answer is given both $(P)$ and $(Q)$ are true. I am stuck here.
If the solutions $y_1$ and $y_2$ are linearly independent then how can we show this.
The converse of this fact is false, and you just found a counterexample to show it is false.
So there is no problem that both $(P)$ and $(Q)$ are true.
The only thing you still need to do is show that $y_1$ and $y_2$ are linearly independent.