What is the solution for 2y^{''}+8y^'+80y=50,y(0)=0,y^'(0)=0 ?

The solution for 2y^{''}+8y^'+80y=50,y(0)=0,y^'(0)=0 is y=e^{-2t}(-5/8 cos(6t)-5/24 sin(6t))+5/8

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2y^{''}+8y^'+80y=50,y(0)=0,y^'(0)=0

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 50, y (0) = 0, y^{^{'}} (0) = 0

y = e^{- 2 t} (- \frac{5}{8} cos (6 t) - \frac{5}{24} sin (6 t)) + \frac{5}{8}

Solution steps

2 y^{''} (t) + 8 y^{'} (t) + 80 y = 50

Solve linear ODE:

y = e^{- 2 t} (- \frac{5}{8} cos (6 t) - \frac{5}{24} sin (6 t)) + \frac{5}{8}

y = e^{- 2 t} (- \frac{5}{8} cos (6 t) - \frac{5}{24} sin (6 t)) + \frac{5}{8}

y^{^{''}} + 4 y = t^{2} + 10 e^{t}, y (0) = 0, y^{^{'}} (0) = 7

y^{^{''}} - 2 y^{^{'}} - 3 y = - 3 e^{- x}

x^{^{''}} + 2 x^{^{'}} + 7 x = 6 cos (2 t)

y^{^{''}} + 36 y = - 36 sin (6 t)

y^{^{'''}} + y^{^{''}} = x

What is the solution for 2y^{''}+8y^'+80y=50,y(0)=0,y^'(0)=0 ?
The solution for 2y^{''}+8y^'+80y=50,y(0)=0,y^'(0)=0 is y=e^{-2t}(-5/8 cos(6t)-5/24 sin(6t))+5/8