What is the solution for x^2y^{''}-2xy^'-4y=0 ?

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

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x^2y^{''}-2xy^'-4y=0

x^{2} y^{^{''}} - 2 x y^{^{'}} - 4 y = 0

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

Solution steps

x^{2} y^{''} (x) - 2 x y^{'} (x) - 4 y = 0

Solve linear ODE:

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

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

\int (\frac{1}{x + 1}) d x

\frac{d}{d x} (\frac{2 a}{x})

\frac{\partial}{\partial x} (7 e^{- 4 x})

\int_{2}^{6} f (x) d x

(2, 3), (4, 8)

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