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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + y = 0, y (0) = 3, y^{^{'}} (0) = 5

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

x^{^{''}} + a x^{^{'}} + b x = 0

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 4, y^{^{'}} (0) = - 2

\frac{d ^{2} y}{d t ^{2}} + 6 \frac{d y}{d t} + 9 y = 0

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