What is the solution for t^2y^{''}+4ty^'-18y=0 ?

The solution for t^2y^{''}+4ty^'-18y=0 is y=c_{1}t^3+(c_{2})/(t^6)

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t^2y^{''}+4ty^'-18y=0

t^{2} y^{^{''}} + 4 t y^{^{'}} - 18 y = 0

y = c_{1} t^{3} + \frac{c _{2}}{t ^{6}}

Solution steps

t^{2} y^{''} (t) + 4 t y^{'} (t) - 18 y = 0

Solve linear ODE:

y = c_{1} t^{3} + \frac{c _{2}}{t ^{6}}

y = c_{1} t^{3} + \frac{c _{2}}{t ^{6}}

\frac{\partial}{\partial x} (- 2 x e^{- 3 x y})

d er i v a t i v e e^{\frac{3}{x}}

\int cos^{3} (\frac{x}{13}) d x

d er i v a t i v e y = (9 x^{4} + 5 x)^{3}

\int e^{8 x + 7} d x

What is the solution for t^2y^{''}+4ty^'-18y=0 ?
The solution for t^2y^{''}+4ty^'-18y=0 is y=c_{1}t^3+(c_{2})/(t^6)