What is the solution for y^{''}+4y^'+7y=0,y(0)=1,y^'(0)=-2 ?

The solution for y^{''}+4y^'+7y=0,y(0)=1,y^'(0)=-2 is y=e^{-2t}cos(sqrt(3)t)

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y^{''}+4y^'+7y=0,y(0)=1,y^'(0)=-2

y^{^{''}} + 4 y^{^{'}} + 7 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

y = e^{- 2 t} cos (3 t)

Solution steps

y^{''} (t) + 4 y^{'} (t) + 7 y = 0

Solve linear ODE:

y = e^{- 2 t} cos (3 t)

y = e^{- 2 t} cos (3 t)

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

y^{^{''}} + 81 y = 0, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} - 7 y^{^{'}} - 8 y = 0, y (0) = 2, y^{^{'}} (0) = 7

y^{^{''}} - y^{^{'}} = - y

4 y^{^{''}} - 36 y^{^{'}} + 81 y = 0

What is the solution for y^{''}+4y^'+7y=0,y(0)=1,y^'(0)=-2 ?
The solution for y^{''}+4y^'+7y=0,y(0)=1,y^'(0)=-2 is y=e^{-2t}cos(sqrt(3)t)