What is the solution for 4/(9.8)y^{''}+0.4y^'+4y=0 ?

The solution for 4/(9.8)y^{''}+0.4y^'+4y=0 is y=e^{-0.49t}(c_{1}cos(3.09190…t)+c_{2}sin(3.09190…t))

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4/(9.8)y^{''}+0.4y^'+4y=0

\frac{4}{9.8} y^{^{''}} + 0.4 y^{^{'}} + 4 y = 0

y = e^{- 0.49 t} (c_{1} cos (3.09190 \dots t) + c_{2} sin (3.09190 \dots t))

Solution steps

\frac{4}{9.8} y^{''} (t) + 0.4 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = e^{- 0.49 t} (c_{1} cos (3.09190 \dots t) + c_{2} sin (3.09190 \dots t))

y = e^{- 0.49 t} (c_{1} cos (3.09190 \dots t) + c_{2} sin (3.09190 \dots t))

(\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y)^{2} = 0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y^{^{'}} (0) = 0, y (0) = 5

100 y^{^{''}} + 125 y^{^{'}} + 100 y = 0

5 y^{^{''}} + 25 y^{^{'}} + 100 y = 0

y^{^{''}} + 49 y = 0, y (0) = 6, y^{^{'}} (0) = - 6

What is the solution for 4/(9.8)y^{''}+0.4y^'+4y=0 ?
The solution for 4/(9.8)y^{''}+0.4y^'+4y=0 is y=e^{-0.49t}(c_{1}cos(3.09190…t)+c_{2}sin(3.09190…t))