What is the solution for y^{''}+2y^'+4y=0,y(0)=0,y^'(0)=1 ?

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

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y^{''}+2y^'+4y=0,y(0)=0,y^'(0)=1

y^{^{''}} + 2 y^{^{'}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{1}{3} e^{- t} sin (3 t)

Solution steps

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

Solve linear ODE:

y = \frac{1}{3} e^{- t} sin (3 t)

y = \frac{1}{3} e^{- t} sin (3 t)

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

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

y^{^{''}} + 36 y = 0, y (\frac{π}{6}) = 1, y^{^{'}} (\frac{π}{6}) = - 2

y^{^{'''}} - y^{^{''}} - 24 y^{^{'}} - 36 y = 0

\frac{d ^{2} y}{d x ^{2}} + 225 y = 0

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