What is the solution for t^2y^{''}+ty^'+9y=0 ?

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

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t^2y^{''}+ty^'+9y=0

t^{2} y^{^{''}} + t y^{^{'}} + 9 y = 0

y = c_{1} cos (3 ln (t)) + c_{2} sin (3 ln (t))

Solution steps

t^{2} y^{''} (t) + t y^{'} (t) + 9 y = 0

Solve linear ODE:

y = c_{1} cos (3 ln (t)) + c_{2} sin (3 ln (t))

y = c_{1} cos (3 ln (t)) + c_{2} sin (3 ln (t))

h \to - 3 lim (\frac{h ^{3} + 8}{h + 2})

\frac{\partial}{\partial x} (csc (x))

y = - \frac{3}{4 x ^{4}} + \frac{24}{5 x ^{5}}

(x + 1) - 1

\frac{d}{d x} (1 + 3 sin^{2} (x))

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