What is the solution for y^{''}+4y^'+20y=0,y(0)=6,y^'(0)=-11 ?

The solution for y^{''}+4y^'+20y=0,y(0)=6,y^'(0)=-11 is y=e^{-2t}(6cos(4t)+1/4 sin(4t))

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y^{''}+4y^'+20y=0,y(0)=6,y^'(0)=-11

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (0) = 6, y^{^{'}} (0) = - 11

y = e^{- 2 t} (6 cos (4 t) + \frac{1}{4} sin (4 t))

Solution steps

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

Solve linear ODE:

y = e^{- 2 t} (6 cos (4 t) + \frac{1}{4} sin (4 t))

y = e^{- 2 t} (6 cos (4 t) + \frac{1}{4} sin (4 t))

y^{^{'''}} + y^{^{''}} - 10 y^{^{'}} + 8 y = 0

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y^{^{'}} = 3, y^{0} (0) = 0

4 y^{^{''}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{'''}} - 7 y^{^{''}} + 25 y^{^{'}} - 39 y = 0

2 \cdot 0.04 \cdot x^{^{'}} + 0.25 \cdot 0.04 \cdot x^{^{''}} + x = 0

What is the solution for y^{''}+4y^'+20y=0,y(0)=6,y^'(0)=-11 ?
The solution for y^{''}+4y^'+20y=0,y(0)=6,y^'(0)=-11 is y=e^{-2t}(6cos(4t)+1/4 sin(4t))