What is the solution for t^2y^{''}+5ty^'-12y=0 ?

The solution for t^2y^{''}+5ty^'-12y=0 is y=c_{1}t^2+(c_{2})/(t^6)

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t^2y^{''}+5ty^'-12y=0

t^{2} y^{^{''}} + 5 t y^{^{'}} - 12 y = 0

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

Solution steps

t^{2} y^{''} (t) + 5 t y^{'} (t) - 12 y = 0

Solve linear ODE:

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

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

\frac{1}{( 1 - x ^{2} ) ^{2}}

10 x - 6 y x^{2} + 1 \frac{d y}{d x} = 0

\frac{d r}{d θ} = \frac{r ^{2} + 5 r θ ^{4}}{θ ^{5}}

y = \frac{2 x}{x ^{2} - 7}, (3, 3)

x, 2 x 1 - x^{2}

What is the solution for t^2y^{''}+5ty^'-12y=0 ?
The solution for t^2y^{''}+5ty^'-12y=0 is y=c_{1}t^2+(c_{2})/(t^6)