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

The solution for y^{''}-(6/x)y^'-(3/(x^2))y=0 is y=c_{1}x^{(7+sqrt(61))/2}+c_{2}x^{(7-sqrt(61))/2}

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

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

y = c_{1} x^{\frac{7 + 61}{2}} + c_{2} x^{\frac{7 - 61}{2}}

Solution steps

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

Solve linear ODE:

y = c_{1} x^{\frac{7 + 61}{2}} + c_{2} x^{\frac{7 - 61}{2}}

y = c_{1} x^{\frac{7 + 61}{2}} + c_{2} x^{\frac{7 - 61}{2}}

n = 1 \sum \infty n - 1

\frac{d}{d x} ((sin (x))^{6})

\frac{d}{d x} (\frac{x ^{2}}{y})

\int \frac{1}{1 - u ^{2}} d u

What is the solution for y^{''}-(6/x)y^'-(3/(x^2))y=0 ?
The solution for y^{''}-(6/x)y^'-(3/(x^2))y=0 is y=c_{1}x^{(7+sqrt(61))/2}+c_{2}x^{(7-sqrt(61))/2}