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I am having trouble finding the general solution of the following second order ODE for $y = y(x)$ without constant coefficients:

$3x^2y'' = 6y$
$x>0$

I realise that it may be possible to simply guess the form of the solution and substitute it back into the the equation but i do not wish to use that approach here.

I would appreciate any help, thanks.

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From the form of this differential equation $$3x^2y'' = 6y$$ it is evident that $y=x^k$ is a potential solution.

Upon substitution of $y=x^k$ we come up with $k=2$ and $k=-1$

Thus the general solution is $$ y= C_1 x^2 +C_2 x^{-1}.$$

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Hint

Simple way $$y''-2\dfrac {y}{x^2} = 0$$ $$x^2y''+2xy'-2xy'-2y= 0$$ $$(x^2y')'-2(xy)' = 0$$ Integrate $$(x^2y')-2(xy) = K_1$$ Divide by $x^4$ $$\dfrac {x^2y'-2xy}{x^4} = \dfrac {K_1}{x^4}$$ $$(\dfrac {y}{x^2})' = \dfrac {K_1}{x^4}$$ integrate again $$\dfrac {y}{x^2} = \int \dfrac {K_1}{x^4}dx +K_2$$ $$\boxed{y(x)=\frac {K_1} x +K_2 x^2}$$

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