(adsbygoogle = window.adsbygoogle || []).push({}); Second order diff.eq. help??

well i am trying to find a solution to this diff. eq, but i get stuck somewhere.

[tex]\ 4x^{2}y''+y=0[/tex]

I first took this substitution

y'=p, y"=p' so the diff. eq becomes of this form

[tex]\ 4x^{2}p'+p=0[/tex]

i think this can be done with the separable of variables thing. so

[tex]\ 4x^{2}\frac{dp}{dx}=-p[/tex]

[tex]\frac{dp}{p}=-\frac{dx}{4x^{2}}[/tex] now i integrate bot sides

[tex] \int\frac{dp}{p}[/tex] = [tex] -\int\frac{dx}{4x^{2}}[/tex], after some calculations i get

[tex] p=A e^{\frac{1}{4x}[/tex], where A is a constant that we get because of integrating bot siedes. NOw i go back to p=y'

so

[tex] y'= A e^{\frac{1}{4x}[/tex], now i think all i need to do is integrate, but i get stuck integrating the right part of this eq. I do not think it has any closed form, right?

but i know that the answer to this equation is in a closed form, that is, it is an elementary function, so where am i going wrong?

I also tried to consider the eq [tex]\ 4x^{2}p'+p=0[/tex] as a linear one, but also when i tried to derive an integrating factor i ended up with a similar expression as [tex] y'= A e^{\frac{1}{4x}[/tex],

so, any hints on how to solve this?

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# Second order diff.eq. help?

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