What is the solution for x^2y^{''}+7xy^'+5y=0 ?

The solution for x^2y^{''}+7xy^'+5y=0 is y=(c_{1})/x+(c_{2})/(x^5)

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x^2y^{''}+7xy^'+5y=0

x^{2} y^{^{''}} + 7 x y^{^{'}} + 5 y = 0

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

Solution steps

x^{2} y^{''} (x) + 7 x y^{'} (x) + 5 y = 0

Solve linear ODE:

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

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

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

\int sin^{18} (x) cos^{3} (x) d x

g (x) = 4 x^{2} (x^{3} + 5 x)

\int (- 8 sec (x) tan (x) + 5 sec^{2} (x)) d x

\frac{d}{d x} ((2 - 7 x^{\frac{3}{2}})^{2})

What is the solution for x^2y^{''}+7xy^'+5y=0 ?
The solution for x^2y^{''}+7xy^'+5y=0 is y=(c_{1})/x+(c_{2})/(x^5)