What is the solution for y^{''}-(3/x)y^'-(6/(x^2))y=0 ?

The solution for y^{''}-(3/x)y^'-(6/(x^2))y=0 is y=c_{1}x^{2+sqrt(10)}+c_{2}x^{2-sqrt(10)}

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y^{''}-(3/x)y^'-(6/(x^2))y=0

y^{^{''}} - (\frac{3}{x}) y^{^{'}} - (\frac{6}{x ^{2}}) y = 0

y = c_{1} x^{2 + 10} + c_{2} x^{2 - 10}

Solution steps

y^{''} (x) - (\frac{3}{x}) y^{'} (x) - (\frac{6}{x ^{2}}) y = 0

Solve linear ODE:

y = c_{1} x^{2 + 10} + c_{2} x^{2 - 10}

y = c_{1} x^{2 + 10} + c_{2} x^{2 - 10}

ln (5 + e^{x y}) = y

\frac{\partial}{\partial x} (ln (2 x + 2 c t))

\int \frac{9 ( csc ( t ) ) ^{2}}{cot ( t )} d t

x \to 0 lim (x^{5} - 6 x^{4} + 7)

\int x 2 x^{2} + 1 d x

What is the solution for y^{''}-(3/x)y^'-(6/(x^2))y=0 ?
The solution for y^{''}-(3/x)y^'-(6/(x^2))y=0 is y=c_{1}x^{2+sqrt(10)}+c_{2}x^{2-sqrt(10)}