What is the solution for 81y^{''}-36y^'+8y=0 ?

The solution for 81y^{''}-36y^'+8y=0 is y=e^{(2t)/9}(c_{1}cos((2t)/9)+c_{2}sin((2t)/9))

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81y^{''}-36y^'+8y=0

81 y^{^{''}} - 36 y^{^{'}} + 8 y = 0

y = e^{\frac{2 t}{9}} (c_{1} cos (\frac{2 t}{9}) + c_{2} sin (\frac{2 t}{9}))

Solution steps

81 y^{''} (t) - 36 y^{'} (t) + 8 y = 0

Solve linear ODE:

y = e^{\frac{2 t}{9}} (c_{1} cos (\frac{2 t}{9}) + c_{2} sin (\frac{2 t}{9}))

y = e^{\frac{2 t}{9}} (c_{1} cos (\frac{2 t}{9}) + c_{2} sin (\frac{2 t}{9}))

\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} - 10 y = 0

y^{^{''}} + 10 y^{^{'}} + 34 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - y = 0, y (1) = 0, y^{^{'}} (1) = - 1

x^{^{''}} - 26 x^{^{'}} + 25 x = 0, x (0) = - 1, x^{^{'}} (0) = 2

D^{2} (D + 4)^{3} y = 0

What is the solution for 81y^{''}-36y^'+8y=0 ?
The solution for 81y^{''}-36y^'+8y=0 is y=e^{(2t)/9}(c_{1}cos((2t)/9)+c_{2}sin((2t)/9))