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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{4 x + 3}{2 x + 1} d x

y = x^{3} - 16 x + 2

\int r^{- 2} d r

x \to 0 + lim (\frac{e ^{x}}{ln ( x )})

\int x^{2} + 2 x - 3 d x

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