What is the solution for 2y^{''}+8y^'+80y=40,y(0)=0,y^'(0)=0 ?

The solution for 2y^{''}+8y^'+80y=40,y(0)=0,y^'(0)=0 is y=e^{-2t}(-1/2 cos(6t)-1/6 sin(6t))+1/2

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2y^{''}+8y^'+80y=40,y(0)=0,y^'(0)=0

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 40, y (0) = 0, y^{^{'}} (0) = 0

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

Solution steps

2 y^{''} (t) + 8 y^{'} (t) + 80 y = 40

Solve linear ODE:

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

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

y^{^{''}} + 16 y = 12 cos (4 x) + 18 sin (4 x)

- \frac{1}{4} y^{^{''}} + y^{^{'}} + y = x^{2} - 2 x

y^{^{''}} - 10 y^{^{'}} + 25 y = \frac{e ^{5 x}}{4 + x ^{2}}

y^{^{''}} - 9 y = sin (3 x)

y^{^{''}} + 4 y = x^{2} + sin (2 x)

What is the solution for 2y^{''}+8y^'+80y=40,y(0)=0,y^'(0)=0 ?
The solution for 2y^{''}+8y^'+80y=40,y(0)=0,y^'(0)=0 is y=e^{-2t}(-1/2 cos(6t)-1/6 sin(6t))+1/2