What is the solution for 2y^{''}+4y^'+3y=0 ?

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

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2y^{''}+4y^'+3y=0

2 y^{^{''}} + 4 y^{^{'}} + 3 y = 0

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

Solution steps

2 y^{''} (t) + 4 y^{'} (t) + 3 y = 0

Solve linear ODE:

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

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

x^{^{''}} - 0.8 x^{^{'}} + 1.16 x = 0

5 y^{^{''}} + 50 y^{^{'}} + 45 y = 0, y (0) = - 8, y^{^{'}} (0) = 0

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

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

9 y^{^{''}} - 6 y^{^{'}} + y = 0, y (0) = 3, y^{^{'}} (0) = 4

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