What is the solution for t^2y^{''}-5ty^'+9y=0 ?

The solution for t^2y^{''}-5ty^'+9y=0 is y=c_{1}t^3+c_{2}ln(t)t^3

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t^2y^{''}-5ty^'+9y=0

t^{2} y^{^{''}} - 5 t y^{^{'}} + 9 y = 0

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

Solution steps

t^{2} y^{''} (t) - 5 t y^{'} (t) + 9 y = 0

Solve linear ODE:

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

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

x cos (3 x)

y^{^{'}} = x e^{y}, y (0) = 0

\frac{d}{d x} (arctan^{2} (\frac{1}{x}))

\frac{d}{d x} (2 sin (x) - 3 cos (x))

\int 20 arctan (x) d x

What is the solution for t^2y^{''}-5ty^'+9y=0 ?
The solution for t^2y^{''}-5ty^'+9y=0 is y=c_{1}t^3+c_{2}ln(t)t^3