What is the solution for t^2y^{''}+19ty^'+162y=0 ?

The solution for t^2y^{''}+19ty^'+162y=0 is y=(c_{1}cos(9ln(t))+c_{2}sin(9ln(t)))/(t^9)

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t^2y^{''}+19ty^'+162y=0

t^{2} y^{^{''}} + 19 t y^{^{'}} + 162 y = 0

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

Solution steps

t^{2} y^{''} (t) + 19 t y^{'} (t) + 162 y = 0

Solve linear ODE:

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

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

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What is the solution for t^2y^{''}+19ty^'+162y=0 ?
The solution for t^2y^{''}+19ty^'+162y=0 is y=(c_{1}cos(9ln(t))+c_{2}sin(9ln(t)))/(t^9)