What is the solution for x^2y^{''}+3xy^'+4y=0 ?

The solution for x^2y^{''}+3xy^'+4y=0 is y=(c_{1}cos(sqrt(3)ln(x))+c_{2}sin(sqrt(3)ln(x)))/x

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x^2y^{''}+3xy^'+4y=0

x^{2} y^{^{''}} + 3 x y^{^{'}} + 4 y = 0

y = \frac{c _{1} cos ( 3 ln ( x ) ) + c _{2} sin ( 3 ln ( x ) )}{x}

Solution steps

x^{2} y^{''} (x) + 3 x y^{'} (x) + 4 y = 0

Solve linear ODE:

y = \frac{c _{1} cos ( 3 ln ( x ) ) + c _{2} sin ( 3 ln ( x ) )}{x}

y = \frac{c _{1} cos ( 3 ln ( x ) ) + c _{2} sin ( 3 ln ( x ) )}{x}

\frac{\partial}{\partial z} (sin (z^{3}) cos (x y))

f (x) = \frac{- 2 x}{x ^{5} - x - 1}

\int yz e^{x y} d x

y = x, (4, 2)

\int \frac{4 x ^{2} + 5 x + 4}{( x ^{2} + 1 ) ^{2}} d x

What is the solution for x^2y^{''}+3xy^'+4y=0 ?
The solution for x^2y^{''}+3xy^'+4y=0 is y=(c_{1}cos(sqrt(3)ln(x))+c_{2}sin(sqrt(3)ln(x)))/x