What is the solution for t^2y^{''}+19ty^'+82y=0 ?

The solution for t^2y^{''}+19ty^'+82y=0 is y=(c_{1}cos(ln(t))+c_{2}sin(ln(t)))/(t^9)

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t^2y^{''}+19ty^'+82y=0

t^{2} y^{^{''}} + 19 t y^{^{'}} + 82 y = 0

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

Solution steps

t^{2} y^{''} (t) + 19 t y^{'} (t) + 82 y = 0

Solve linear ODE:

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

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

(9, - 7), (4, - 8)

\int_{0}^{9} \frac{1}{81 - x ^{2}} d x

\frac{\partial}{\partial x} (6 x y - 5 x^{2} y)

x, 3

\frac{d y}{d x}, x^{2} = \frac{( 3 x + 5 y )}{3 x - 5 y}

What is the solution for t^2y^{''}+19ty^'+82y=0 ?
The solution for t^2y^{''}+19ty^'+82y=0 is y=(c_{1}cos(ln(t))+c_{2}sin(ln(t)))/(t^9)