What is the solution for t^2y^{''}+2ty^'-6y=0 ?

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

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t^2y^{''}+2ty^'-6y=0

t^{2} y^{^{''}} + 2 t y^{^{'}} - 6 y = 0

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

Solution steps

t^{2} y^{''} (t) + 2 t y^{'} (t) - 6 y = 0

Solve linear ODE:

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

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

\int \frac{1}{3 + t} d t

y^{^{'}} = 3 y x^{2} - 3 x^{2}

x \to - 3 lim (\frac{2}{x ^{2}})

\int - 64 (1 + 4 t)^{- 3} d t

e^{x} - e^{- x}

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