What is the solution for t^2y^{''}+19ty^'+130y=0 ?

The solution for t^2y^{''}+19ty^'+130y=0 is y=(c_{1}cos(7ln(t))+c_{2}sin(7ln(t)))/(t^9)

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t^2y^{''}+19ty^'+130y=0

t^{2} y^{^{''}} + 19 t y^{^{'}} + 130 y = 0

y = \frac{c _{1} cos ( 7 ln ( t ) ) + c _{2} sin ( 7 ln ( t ) )}{t ^{9}}

Solution steps

t^{2} y^{''} (t) + 19 t y^{'} (t) + 130 y = 0

Solve linear ODE:

y = \frac{c _{1} cos ( 7 ln ( t ) ) + c _{2} sin ( 7 ln ( t ) )}{t ^{9}}

y = \frac{c _{1} cos ( 7 ln ( t ) ) + c _{2} sin ( 7 ln ( t ) )}{t ^{9}}

\frac{\partial}{\partial x} (e^{- xx} sin (3 x + 1))

\frac{d}{d x} (\frac{- x ^{2}}{x + 1})

n = 1 \sum \infty \frac{( n + 6 )}{2}

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

2 x, \frac{x}{2}

What is the solution for t^2y^{''}+19ty^'+130y=0 ?
The solution for t^2y^{''}+19ty^'+130y=0 is y=(c_{1}cos(7ln(t))+c_{2}sin(7ln(t)))/(t^9)