What is the solution for 81y^{''}+36y^'+4y=0 ?

The solution for 81y^{''}+36y^'+4y=0 is y=c_{1}e^{-(2t)/9}+c_{2}e^{-(2t)/9}t

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81y^{''}+36y^'+4y=0

81 y^{^{''}} + 36 y^{^{'}} + 4 y = 0

y = c_{1} e^{- \frac{2 t}{9}} + c_{2} e^{- \frac{2 t}{9}} t

Solution steps

81 y^{''} (t) + 36 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{2 t}{9}} + c_{2} e^{- \frac{2 t}{9}} t

y = c_{1} e^{- \frac{2 t}{9}} + c_{2} e^{- \frac{2 t}{9}} t

y \cdot y^{^{''}} + (y)^{2} = 0, y (0) = 4, y^{^{'}} (0) = 6

y^{^{''}} + y^{^{'}} - 20 y = 0, y (0) = - 3, y^{^{'}} (0) = 33

y^{^{''}} - y^{^{'}} + 2 y = 0, y (0) = 5

x^{^{''}} + c^{2} \cdot x = 0

y^{^{'''}} - 8 y^{^{''}} - y^{^{'}} + 8 y = 0

What is the solution for 81y^{''}+36y^'+4y=0 ?
The solution for 81y^{''}+36y^'+4y=0 is y=c_{1}e^{-(2t)/9}+c_{2}e^{-(2t)/9}t