What is the solution for 2x^{''}+16x^'+64x=0,x(0)=0,x^'(0)=0.2 ?

The solution for 2x^{''}+16x^'+64x=0,x(0)=0,x^'(0)=0.2 is x=(0.2)/4 e^{-4t}sin(4t)

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2x^{''}+16x^'+64x=0,x(0)=0,x^'(0)=0.2

2 x^{^{''}} + 16 x^{^{'}} + 64 x = 0, x (0) = 0, x^{^{'}} (0) = 0.2

x = \frac{0.2}{4} e^{- 4 t} sin (4 t)

Solution steps

2 x^{''} (t) + 16 x^{'} (t) + 64 x = 0

Solve linear ODE:

x = \frac{0.2}{4} e^{- 4 t} sin (4 t)

x = \frac{0.2}{4} e^{- 4 t} sin (4 t)

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

y^{^{'''}} + 8 y^{^{''}} + 19 y^{^{'}} + 12 y = 0

y^{^{''}} + 25 y = 0, y (0) = 0, y^{^{'}} (0) = 4

81 y^{^{''}} + 108 y^{^{'}} - 13 y = 0, y (0) = 1, y^{^{'}} (0) = 0

x^{^{''}} + 0.5 x^{^{'}} + 4 x = 0

What is the solution for 2x^{''}+16x^'+64x=0,x(0)=0,x^'(0)=0.2 ?
The solution for 2x^{''}+16x^'+64x=0,x(0)=0,x^'(0)=0.2 is x=(0.2)/4 e^{-4t}sin(4t)