What is the solution for 2x^2y^{''}+6xy^'+2y=0 ?

The solution for 2x^2y^{''}+6xy^'+2y=0 is y=(c_{1})/x+(c_{2}ln(x))/x

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2x^2y^{''}+6xy^'+2y=0

2 x^{2} y^{^{''}} + 6 x y^{^{'}} + 2 y = 0

y = \frac{c _{1}}{x} + \frac{c _{2} ln ( x )}{x}

Solution steps

2 x^{2} y^{''} (x) + 6 x y^{'} (x) + 2 y = 0

Solve linear ODE:

y = \frac{c _{1}}{x} + \frac{c _{2} ln ( x )}{x}

y = \frac{c _{1}}{x} + \frac{c _{2} ln ( x )}{x}

\frac{\partial}{\partial x} (x^{3} + x)

\int_{b}^{1} 3 ln (x) d x

\int 5 x \frac{3}{2} d x

\frac{d y}{d x} - (sin (x)) y = 2 sin (x)

\frac{d}{d x} (\frac{e ^{- x}}{x})

What is the solution for 2x^2y^{''}+6xy^'+2y=0 ?
The solution for 2x^2y^{''}+6xy^'+2y=0 is y=(c_{1})/x+(c_{2}ln(x))/x