What is the solution for y^{''}+4y^'+6y=0,y(0)=2,y^'(0)=4 ?

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

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y^{''}+4y^'+6y=0,y(0)=2,y^'(0)=4

y^{^{''}} + 4 y^{^{'}} + 6 y = 0, y (0) = 2, y^{^{'}} (0) = 4

y = e^{- 2 t} (2 cos (2 t) + 42 sin (2 t))

Solution steps

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

Solve linear ODE:

y = e^{- 2 t} (2 cos (2 t) + 42 sin (2 t))

y = e^{- 2 t} (2 cos (2 t) + 42 sin (2 t))

y^{^{''}} + 6 y^{^{'}} - 216 y = 0

y^{^{''}} + 6 y^{^{'}} + 16 y = 0

5 y^{^{''}} + 20 y = 0

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

4 y^{^{''}} - 8 y^{^{'}} + 3 y = 0, y (0) = 2, y^{^{'}} (0) = 1

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