What is the solution for t^2y^{''}+19ty^'+90y=0 ?

The solution for t^2y^{''}+19ty^'+90y=0 is y=(c_{1}cos(3ln(t))+c_{2}sin(3ln(t)))/(t^9)

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t^2y^{''}+19ty^'+90y=0

t^{2} y^{^{''}} + 19 t y^{^{'}} + 90 y = 0

y = \frac{c _{1} cos ( 3 ln ( t ) ) + c _{2} sin ( 3 ln ( t ) )}{t ^{9}}

Solution steps

t^{2} y^{''} (t) + 19 t y^{'} (t) + 90 y = 0

Solve linear ODE:

y = \frac{c _{1} cos ( 3 ln ( t ) ) + c _{2} sin ( 3 ln ( t ) )}{t ^{9}}

y = \frac{c _{1} cos ( 3 ln ( t ) ) + c _{2} sin ( 3 ln ( t ) )}{t ^{9}}

- 2 y^{^{''}} + 8 y = 0

e^{\frac{t}{9}}

f (x) = (3 x^{2} - 2 x + 1)^{\frac{1}{2}}

\int 4 cos (\frac{nπ x}{4}) d x

\frac{x}{x ^{2} + 0.0184}

What is the solution for t^2y^{''}+19ty^'+90y=0 ?
The solution for t^2y^{''}+19ty^'+90y=0 is y=(c_{1}cos(3ln(t))+c_{2}sin(3ln(t)))/(t^9)