What is the solution for 2y^{''}+8y^'+80y=0,y(0)=0 ?

The solution for 2y^{''}+8y^'+80y=0,y(0)=0 is y=c_{2}e^{-2t}sin(6t)

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2y^{''}+8y^'+80y=0,y(0)=0

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 0, y (0) = 0

y = c_{2} e^{- 2 t} sin (6 t)

Solution steps

2 y^{''} (t) + 8 y^{'} (t) + 80 y = 0

Solve linear ODE:

y = c_{2} e^{- 2 t} sin (6 t)

y = c_{2} e^{- 2 t} sin (6 t)

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = 0, y^{^{'}} (0) = 4

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} - 8 x = 0

8 x^{^{''}} + x = 0

1.5 x^{^{''}} + 12 x^{^{'}} + 24 x = 0

y^{^{'''}} + 6 y^{^{''}} - y^{^{'}} - 3 y = 0

What is the solution for 2y^{''}+8y^'+80y=0,y(0)=0 ?
The solution for 2y^{''}+8y^'+80y=0,y(0)=0 is y=c_{2}e^{-2t}sin(6t)