Frobenius method on $y''+x^2.y=0\; ; \; y(1)=0 \; ; \; y'(1)=1$ ($y$ roots)

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The differential problem is $$y''+x^2.y=0$$ $$y(1)=0\; ;\; y'(1)=1$$ I have to find a possible solution for this problem, show that $y(x)$ has another roots and find the next root for $x>1$.

Using the Frobenius method, I found that $$y(x)=\frac{1}{4}.\sum_{n=1}^\infty \frac{(-1)^n}{n!.(4.n-1).(4.n-5).\cdots. 3}.a_0.x^{4.n}+a_0+\frac{1}{4}.\sum_{n=1}^\infty\frac{(-1)^n}{n!.(4.n+1).(4.n-3).\cdots.5}.a_1.x^{4.n+1}+a_1$$

How can I discover the values of $a_0$ and $a_1$ by using the expressions $y(1)=0\; ;\; y'(1)=1$ ?