What is the solution for y^{''}+3/x y^'-8/(x^2)y=0 ?

The solution for y^{''}+3/x y^'-8/(x^2)y=0 is y=c_{1}x^2+(c_{2})/(x^4)

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y^{''}+3/x y^'-8/(x^2)y=0

y^{^{''}} + \frac{3}{x} y^{^{'}} - \frac{8}{x ^{2}} y = 0

y = c_{1} x^{2} + \frac{c _{2}}{x ^{4}}

Solution steps

y^{''} (x) + \frac{3}{x} y^{'} (x) - \frac{8}{x ^{2}} y = 0

Solve linear ODE:

y = c_{1} x^{2} + \frac{c _{2}}{x ^{4}}

y = c_{1} x^{2} + \frac{c _{2}}{x ^{4}}

\int_{- 5}^{5} \frac{1}{x} d x

y^{^{''}} + 10 y^{^{'}} + 25 y = t^{- 2} e^{- 5 t}

\frac{\partial}{\partial x} (e^{2 x} cos (3 y))

\frac{d}{d x} (x (4 x - 5)^{3})

\int_{0}^{2} 4 x e^{x} d x

What is the solution for y^{''}+3/x y^'-8/(x^2)y=0 ?
The solution for y^{''}+3/x y^'-8/(x^2)y=0 is y=c_{1}x^2+(c_{2})/(x^4)