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

The solution for t^2y^{''}+2ty^'-12y=0 is y=c_{1}t^3+(c_{2})/(t^4)

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

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

y = c_{1} t^{3} + \frac{c _{2}}{t ^{4}}

Solution steps

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

Solve linear ODE:

y = c_{1} t^{3} + \frac{c _{2}}{t ^{4}}

y = c_{1} t^{3} + \frac{c _{2}}{t ^{4}}

f (x) = 9 x^{\frac{7}{5}} - 5 x^{2} + 1 0^{4}

3 x + sin (5 x)

y = \frac{x}{4 x + 7}

f (x) = ln (cos (x))

\frac{d}{d x} (9 e^{x})

What is the solution for t^2y^{''}+2ty^'-12y=0 ?
The solution for t^2y^{''}+2ty^'-12y=0 is y=c_{1}t^3+(c_{2})/(t^4)