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

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

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

x^{2} y^{^{''}} + 3 x y^{^{'}} + 3 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) + 3 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}

\frac{\partial}{\partial t} (4 t)

f (x) = 3 x + 8

n = 0 \sum \infty ln (\frac{1}{3 ^{n}})

\int_{\frac{1}{3}}^{7} 16 x ln (3 x) d x

\frac{\partial}{\partial x} (arcsin (- 2 x^{4}))

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