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)}

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

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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))}