What is the solution for t^2y^{''}+ty^'+64y=0 ?

The solution for t^2y^{''}+ty^'+64y=0 is y=c_{1}cos(8ln(t))+c_{2}sin(8ln(t))

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t^2y^{''}+ty^'+64y=0

t^{2} y^{^{''}} + t y^{^{'}} + 64 y = 0

y = c_{1} cos (8 ln (t)) + c_{2} sin (8 ln (t))

Solution steps

t^{2} y^{''} (t) + t y^{'} (t) + 64 y = 0

Solve linear ODE:

y = c_{1} cos (8 ln (t)) + c_{2} sin (8 ln (t))

y = c_{1} cos (8 ln (t)) + c_{2} sin (8 ln (t))

\frac{d}{d v} (\frac{m}{v})

1 - 2 x

t \to 1 + lim (\frac{4}{( t - 1 ) ^{3}})

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

\frac{d}{d x} (x^{8 e^{5 x}})

What is the solution for t^2y^{''}+ty^'+64y=0 ?
The solution for t^2y^{''}+ty^'+64y=0 is y=c_{1}cos(8ln(t))+c_{2}sin(8ln(t))