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

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

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y^{''}-3/x y^'-4/(x^2)y=0

y^{^{''}} - \frac{3}{x} y^{^{'}} - \frac{4}{x ^{2}} y = 0

y = c_{1} x^{2 (1 + 2)} + c_{2} x^{2 (1 - 2)}

Solution steps

y^{''} (x) - \frac{3}{x} y^{'} (x) - \frac{4}{x ^{2}} y = 0

Solve linear ODE:

y = c_{1} x^{2 (1 + 2)} + c_{2} x^{2 (1 - 2)}

y = c_{1} x^{2 (1 + 2)} + c_{2} x^{2 (1 - 2)}

\frac{\partial}{\partial x} (17 x^{3} y^{2})

\frac{d}{d x} (e^{x} (5 x^{2} - 2 x + 3))

\frac{d y}{d x} = \frac{y sin ( x )}{1 - 2 y}

y^{^{'}} = e^{y - x}, y (0) = 0

\frac{d}{d x} (c x^{- 8})

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