What is the solution for t^2y^{''}+11ty^'+25y=0 ?

The solution for t^2y^{''}+11ty^'+25y=0 is y=(c_{1})/(t^5)+(c_{2}ln(t))/(t^5)

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t^2y^{''}+11ty^'+25y=0

t^{2} y^{^{''}} + 11 t y^{^{'}} + 25 y = 0

y = \frac{c _{1}}{t ^{5}} + \frac{c _{2} ln ( t )}{t ^{5}}

Solution steps

t^{2} y^{''} (t) + 11 t y^{'} (t) + 25 y = 0

Solve linear ODE:

y = \frac{c _{1}}{t ^{5}} + \frac{c _{2} ln ( t )}{t ^{5}}

y = \frac{c _{1}}{t ^{5}} + \frac{c _{2} ln ( t )}{t ^{5}}

\frac{\partial}{\partial x} (e^{- 5 x^{2} - 2 y^{2}})

\int \frac{t ^{3} + 2}{t ^{4} + 8 t - 3} d t

x \to \infty - lim (e^{x})

\int_{0}^{x} e^{x + y} d y

\frac{x + 1}{x + 2}

What is the solution for t^2y^{''}+11ty^'+25y=0 ?
The solution for t^2y^{''}+11ty^'+25y=0 is y=(c_{1})/(t^5)+(c_{2}ln(t))/(t^5)