What is the solution for y^{''}+4y^'+6y=0,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+4y^'+6y=0,y(0)=1,y^'(0)=0 is y=e^{-2t}(cos(sqrt(2)t)+sqrt(2)sin(sqrt(2)t))

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y^{''}+4y^'+6y=0,y(0)=1,y^'(0)=0

y^{^{''}} + 4 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 6 y = 0

Solve linear ODE:

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

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

y^{^{'}} = x + y, y (0) = 9

\frac{d}{d x} (x + 1 - tan (\frac{x}{4}))

x \to 6 - lim (\frac{1}{x - 6})

\int \frac{9}{( x - 4 ) ( x + 2 )} d x

\int t^{4} e^{t^{5}} d t

What is the solution for y^{''}+4y^'+6y=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+4y^'+6y=0,y(0)=1,y^'(0)=0 is y=e^{-2t}(cos(sqrt(2)t)+sqrt(2)sin(sqrt(2)t))