What is the solution for t^2y^{''}+13ty^'+36y=0 ?

The solution for t^2y^{''}+13ty^'+36y=0 is y=(c_{1})/(t^6)+(c_{2}ln(t))/(t^6)

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t^2y^{''}+13ty^'+36y=0

t^{2} y^{^{''}} + 13 t y^{^{'}} + 36 y = 0

y = \frac{c _{1}}{t ^{6}} + \frac{c _{2} ln ( t )}{t ^{6}}

Solution steps

t^{2} y^{''} (t) + 13 t y^{'} (t) + 36 y = 0

Solve linear ODE:

y = \frac{c _{1}}{t ^{6}} + \frac{c _{2} ln ( t )}{t ^{6}}

y = \frac{c _{1}}{t ^{6}} + \frac{c _{2} ln ( t )}{t ^{6}}

9 y^{^{''}} - 60 y^{^{'}} + 51 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\frac{d}{d x} ((sin (2 x) + 1)^{2})

y = x^{2} - 1, (- 2, 3)

f (x) = 4 x^{2} - 4 x + 1, at x = 4

\frac{\partial}{\partial x} (y \cdot e^{x})

What is the solution for t^2y^{''}+13ty^'+36y=0 ?
The solution for t^2y^{''}+13ty^'+36y=0 is y=(c_{1})/(t^6)+(c_{2}ln(t))/(t^6)