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})

\int 2 x (x^{2} + 15)^{7} d x

\frac{d}{d x} (5 + cos (x))

y^{^{'}} + 2 y = x y^{2}

f (x) = (2 x - 8)^{4} (x^{2} + x + 1)^{5}

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)