What is the solution for t^2y^{''}+2ty^'-30y=0 ?

The solution for t^2y^{''}+2ty^'-30y=0 is y=c_{1}t^5+(c_{2})/(t^6)

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t^2y^{''}+2ty^'-30y=0

t^{2} y^{^{''}} + 2 t y^{^{'}} - 30 y = 0

y = c_{1} t^{5} + \frac{c _{2}}{t ^{6}}

Solution steps

t^{2} y^{''} (t) + 2 t y^{'} (t) - 30 y = 0

Solve linear ODE:

y = c_{1} t^{5} + \frac{c _{2}}{t ^{6}}

y = c_{1} t^{5} + \frac{c _{2}}{t ^{6}}

x \to 3 lim (\frac{( x + 3 )}{x - 3})

\int (ln (5 x))^{2} d x

\int \frac{( 17 x + 3 )}{( 4 x + 1 ) ( x - 1 )} d x

x \to 0 lim (x^{2} ln (x))

\frac{d}{d x} (\frac{1}{5 sin ( x )} + \frac{csc ( x )}{4})

What is the solution for t^2y^{''}+2ty^'-30y=0 ?
The solution for t^2y^{''}+2ty^'-30y=0 is y=c_{1}t^5+(c_{2})/(t^6)