I was working on following problem:
Let $y_1$ and $y_2$ be solutions of
$$x^2y'' + y' + (\sin x)y = 0$$
satisfying
$$y_1(0) = 0, y_1'(0)=1,y_2(0) = 1, y_2'(0)=0 $$.
I worked like following:
since wronskian $W$ is given as $$W = ce^{-\int-\frac{1}{x^2}\,dx}~,$$ wronskian is not zero except the point $x=0$. Thus, The two given solutions are L.I. and hence $y_1$ and $y_2$ do not have common zeroes. Am I right in concluding this? Kindly rectify if somehwere I went wrong. Thanks for giving time.
The question looks incomplete and unanswered for a long time: but guessing from OP's attempt, I think (s)he is trying to find the Wronskian.
Here's couple of hints that will help you solve the question:
$y_1(0) = 0, y_1'(0)=1,y_2(0) = 1, y_2'(0)=0 $. What does that tell you about $W_{(y_1,y_2)}(0)$ ($W_{(y_1,y_2)}(x)=\left|\begin{matrix}y_1(x) &&y_2(x)\\y'_1(x) && y'_2(x)\end{matrix}\right|$)?
A little $\color{red}{\text{correction}}$ and $\color{green}{\text{added details}}$ into Abel's identity which you seem to have used in your attempt:
$$W(x)=\color{green}{W(0)}\times \color{red}{e^{-\int_0^x \frac1{x^2}dx}}=?$$