What is the solution for 16y^{''}+8y^'+145y=0 ?

The solution for 16y^{''}+8y^'+145y=0 is y=e^{-t/4}(c_{1}cos(3t)+c_{2}sin(3t))

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16y^{''}+8y^'+145y=0

16 y^{^{''}} + 8 y^{^{'}} + 145 y = 0

y = e^{- \frac{t}{4}} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

16 y^{''} (t) + 8 y^{'} (t) + 145 y = 0

Solve linear ODE:

y = e^{- \frac{t}{4}} (c_{1} cos (3 t) + c_{2} sin (3 t))

y = e^{- \frac{t}{4}} (c_{1} cos (3 t) + c_{2} sin (3 t))

2 \frac{d ^{2} y}{d x ^{2}} + 20 \frac{d y}{d x} + 51 y = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y^{^{'}} (0) = 1, y (0) = 1

y^{^{'''}} + 3 y^{^{''}} - 4 y = 0, y (1) = \frac{1}{2}

9 y^{^{''}} - 12 y^{^{'}} + 40 y = 0, y (0) = 8, y^{^{'}} (0) = 3

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 4, y^{^{'}} (0) = 10

What is the solution for 16y^{''}+8y^'+145y=0 ?
The solution for 16y^{''}+8y^'+145y=0 is y=e^{-t/4}(c_{1}cos(3t)+c_{2}sin(3t))