What is the solution for x^2y^{''}+2xy^'+100.25y=0 ?

The solution for x^2y^{''}+2xy^'+100.25y=0 is y=(c_{1}cos(10ln(x))+c_{2}sin(10ln(x)))/(sqrt(x))

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x^2y^{''}+2xy^'+100.25y=0

x^{2} y^{^{''}} + 2 x y^{^{'}} + 100.25 y = 0

y = \frac{c _{1} cos ( 10 ln ( x ) ) + c _{2} sin ( 10 ln ( x ) )}{x}

Solution steps

x^{2} y^{''} (x) + 2 x y^{'} (x) + 100.25 y = 0

Solve linear ODE:

y = \frac{c _{1} cos ( 10 ln ( x ) ) + c _{2} sin ( 10 ln ( x ) )}{x}

y = \frac{c _{1} cos ( 10 ln ( x ) ) + c _{2} sin ( 10 ln ( x ) )}{x}

\frac{d}{d x} (\frac{150 x}{x + 8})

f (x) = (ln (x))^{2}

f (x) = - tan^{2} (x)

y^{^{''}} - 4 y = e^{2 t}

3 x^{2} ln (7 x)

What is the solution for x^2y^{''}+2xy^'+100.25y=0 ?
The solution for x^2y^{''}+2xy^'+100.25y=0 is y=(c_{1}cos(10ln(x))+c_{2}sin(10ln(x)))/(sqrt(x))