What is the solution for 2y^{''}+4y^'+20y=0 ?

The solution for 2y^{''}+4y^'+20y=0 is y=e^{-t}(c_{1}cos(3t)+c_{2}sin(3t))

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2y^{''}+4y^'+20y=0

2 y^{^{''}} + 4 y^{^{'}} + 20 y = 0

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

Solution steps

2 y^{''} (t) + 4 y^{'} (t) + 20 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} - 96 y = 0, y (0) = 13, y^{^{'}} (0) = 36

y^{^{''}} + 25 y^{^{'}} + \frac{1}{0.046} y = 0

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

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

y^{^{''}} - 3 y^{^{'}} = 0, y (0) = 2 - 2 e^{3}, y (1) = 0

What is the solution for 2y^{''}+4y^'+20y=0 ?
The solution for 2y^{''}+4y^'+20y=0 is y=e^{-t}(c_{1}cos(3t)+c_{2}sin(3t))