What is the solution for 2y^{''}+8y^'+80y=1,y(0)=0,y^'(0)=0 ?

The solution for 2y^{''}+8y^'+80y=1,y(0)=0,y^'(0)=0 is y=e^{-2t}(-1/80 cos(6t)-1/240 sin(6t))+1/80

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2y^{''}+8y^'+80y=1,y(0)=0,y^'(0)=0

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 1, y (0) = 0, y^{^{'}} (0) = 0

y = e^{- 2 t} (- \frac{1}{80} cos (6 t) - \frac{1}{240} sin (6 t)) + \frac{1}{80}

Solution steps

2 y^{''} (t) + 8 y^{'} (t) + 80 y = 1

Solve linear ODE:

y = e^{- 2 t} (- \frac{1}{80} cos (6 t) - \frac{1}{240} sin (6 t)) + \frac{1}{80}

y = e^{- 2 t} (- \frac{1}{80} cos (6 t) - \frac{1}{240} sin (6 t)) + \frac{1}{80}

D (x) = 82.5 - 2 \cdot 5 x

y^{^{''}} + 6 y = 8 t

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3} e^{2 t}, y^{^{'}} (0) = 0

y^{^{''}} - y^{^{'}} - 6 y = 12 cos (3 x)

x^{^{''}} + 5 x^{^{'}} + 6 x = 10

What is the solution for 2y^{''}+8y^'+80y=1,y(0)=0,y^'(0)=0 ?
The solution for 2y^{''}+8y^'+80y=1,y(0)=0,y^'(0)=0 is y=e^{-2t}(-1/80 cos(6t)-1/240 sin(6t))+1/80