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

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

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

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

y = c_{1} cos (2 ln (x)) + c_{2} sin (2 ln (x))

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 ln (x)) + c_{2} sin (2 ln (x))

y = c_{1} cos (2 ln (x)) + c_{2} sin (2 ln (x))

\frac{\partial}{\partial y} (3 x^{2} + y)

\int (sec^{3} (x)) d x

f (x) = x^{2} - 5 x + 5, (0, 5)

\frac{d}{d t} (e^{t} (cos (t)))

\frac{d}{d x} (x (ln (x) - 1))

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