What is the solution for y^{''}+4y^'+20y=0,y(0)=2,y^'(0)=-8 ?

The solution for y^{''}+4y^'+20y=0,y(0)=2,y^'(0)=-8 is y=e^{-2t}(2cos(4t)-sin(4t))

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y^{''}+4y^'+20y=0,y(0)=2,y^'(0)=-8

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = 2, y^{^{'}} (0) = - 8

y = e^{- 2 t} (2 cos (4 t) - sin (4 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = e^{- 2 t} (2 cos (4 t) - sin (4 t))

y = e^{- 2 t} (2 cos (4 t) - sin (4 t))

(y^{^{''}} - 6 y^{^{'}} + 9 y)^{3} = 0

y^{^{''''}} + 4 y^{^{'''}} - 3 y^{^{''}} - 3 y^{^{'}} + \frac{1}{2} y = 0

(d^{3} - 6 d^{2} + 11 d - 6) y = 0

8 y^{^{''''}} + 6 y^{^{'''}} + 33 y^{^{''}} + 24 y^{^{'}} + 4 y = 0

y^{^{''}} - 14 y^{^{'}} - 49 y = 0

What is the solution for y^{''}+4y^'+20y=0,y(0)=2,y^'(0)=-8 ?
The solution for y^{''}+4y^'+20y=0,y(0)=2,y^'(0)=-8 is y=e^{-2t}(2cos(4t)-sin(4t))