What is the solution for y^{''}-(3/x)y^'-(5/(x^2))y=0 ?

The solution for y^{''}-(3/x)y^'-(5/(x^2))y=0 is y=c_{1}x^5+(c_{2})/x

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y^{''}-(3/x)y^'-(5/(x^2))y=0

y^{^{''}} - (\frac{3}{x}) y^{^{'}} - (\frac{5}{x ^{2}}) y = 0

y = c_{1} x^{5} + \frac{c _{2}}{x}

Solution steps

y^{''} (x) - (\frac{3}{x}) y^{'} (x) - (\frac{5}{x ^{2}}) y = 0

Solve linear ODE:

y = c_{1} x^{5} + \frac{c _{2}}{x}

y = c_{1} x^{5} + \frac{c _{2}}{x}

\int \frac{25}{( x - 3 ) ( x ^{2} + 16 )} d x

\int e^{4 x} cos (6 x) d x

x^{2} - 4 x

\frac{d x}{d t} = 3 x t^{2}

3 sin (2 t)

What is the solution for y^{''}-(3/x)y^'-(5/(x^2))y=0 ?
The solution for y^{''}-(3/x)y^'-(5/(x^2))y=0 is y=c_{1}x^5+(c_{2})/x