What is the solution for t^2y^{''}+19ty^'+145y=0 ?

The solution for t^2y^{''}+19ty^'+145y=0 is y=(c_{1}cos(8ln(t))+c_{2}sin(8ln(t)))/(t^9)

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t^2y^{''}+19ty^'+145y=0

t^{2} y^{^{''}} + 19 t y^{^{'}} + 145 y = 0

y = \frac{c _{1} cos ( 8 ln ( t ) ) + c _{2} sin ( 8 ln ( t ) )}{t ^{9}}

Solution steps

t^{2} y^{''} (t) + 19 t y^{'} (t) + 145 y = 0

Solve linear ODE:

y = \frac{c _{1} cos ( 8 ln ( t ) ) + c _{2} sin ( 8 ln ( t ) )}{t ^{9}}

y = \frac{c _{1} cos ( 8 ln ( t ) ) + c _{2} sin ( 8 ln ( t ) )}{t ^{9}}

\int \frac{x}{x ^{2} + 8} d x

\frac{d}{d x} (x^{8} arctan (x^{3}))

y^{^{'}} = k (y - 70)

\int \frac{ln ( 8 x )}{x} d x

\frac{\partial}{\partial z} (e^{yz})

What is the solution for t^2y^{''}+19ty^'+145y=0 ?
The solution for t^2y^{''}+19ty^'+145y=0 is y=(c_{1}cos(8ln(t))+c_{2}sin(8ln(t)))/(t^9)