What is the solution for x^2y^{''}+3xy^'+5y=0 ?

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

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x^2y^{''}+3xy^'+5y=0

x^{2} y^{^{''}} + 3 x y^{^{'}} + 5 y = 0

y = \frac{c _{1} cos ( 2 ln ( x ) ) + c _{2} sin ( 2 ln ( x ) )}{x}

Solution steps

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

Solve linear ODE:

y = \frac{c _{1} cos ( 2 ln ( x ) ) + c _{2} sin ( 2 ln ( x ) )}{x}

y = \frac{c _{1} cos ( 2 ln ( x ) ) + c _{2} sin ( 2 ln ( x ) )}{x}

\int sin^{5} (4 x) cos (4 x) d x

\frac{\partial}{\partial y} (x e^{x^{3}})

y = 2 x - 1

x \to 7 lim (\frac{π x}{6})

\int \frac{5 x}{( x - 2 ) ^{2}} d x

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