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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 0 - lim (e^{2 x})

\frac{d y}{d x} = y^{2} sin (x)

\frac{d r}{d t} = - r^{3}

1600 y^{^{''}} - 9 y = 0, y (0) = 7, y^{^{'}} (0) = 6

x \to - 1 lim (x^{2} + 4 x)

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