What is the solution for t^2y^{''}+3ty^'-8y=0 ?

The solution for t^2y^{''}+3ty^'-8y=0 is y=c_{1}t^2+(c_{2})/(t^4)

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t^2y^{''}+3ty^'-8y=0

t^{2} y^{^{''}} + 3 t y^{^{'}} - 8 y = 0

y = c_{1} t^{2} + \frac{c _{2}}{t ^{4}}

Solution steps

t^{2} y^{''} (t) + 3 t y^{'} (t) - 8 y = 0

Solve linear ODE:

y = c_{1} t^{2} + \frac{c _{2}}{t ^{4}}

y = c_{1} t^{2} + \frac{c _{2}}{t ^{4}}

\frac{\partial}{\partial y} (x^{4} + x^{2} y)

\frac{d}{d x} (\frac{x ^{2} + 3 x}{6})

\int_{0}^{16} x - 4 d x

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = 4, y^{^{'}} (0) = - 2

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

What is the solution for t^2y^{''}+3ty^'-8y=0 ?
The solution for t^2y^{''}+3ty^'-8y=0 is y=c_{1}t^2+(c_{2})/(t^4)