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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

4 y^{^{''}} + 16 y^{^{'}} + \frac{1}{16} y = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = 0, y (0) = 10, y^{^{'}} (0) = 0

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

y^{^{''}} + 49 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} - 4 y + 3 y = 0, y (0) = - 1, y^{^{'}} (0) = - 5

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