Power series solution of $x^2y'=y$

Question

Not able to understand its solution. Its final answer is $Y=ce^{-1/x}$. I have tried but no able to reach the solution.

Answer 1

$x^2 y' = y \to \dfrac{y'}{y}=\dfrac{1}{x^2}\to\dfrac{dy}{y}=\dfrac{dx}{x^2}$ (Note that $y'=\dfrac{dy}{dx}$)

We now integrate from both sides:

$\int{\dfrac{dy}{y}}=\int{\dfrac{dx}{x^2}}\to ln(|y|)=\dfrac{x^{-1}}{-1}+c$

Take an exponent from both sides:

$|y|=e^{-1/x+c}=e^{-1/x}e^{c}=ce^{-1/x}$

(Note that because $c$ is undetermined, there is no difference between $e^c$ or $c$)

Finally, the equation will be as below where c is a real number:

$y=ce^{-1/x}$