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}

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