What is the solution for 0.5y^{''}+6y^'+50y=24sin(10t) ?

The solution for 0.5y^{''}+6y^'+50y=24sin(10t) is y=e^{-6t}(c_{1}cos(8t)+c_{2}sin(8t))-2/5 cos(10t)

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0.5y^{''}+6y^'+50y=24sin(10t)

0.5 y^{^{''}} + 6 y^{^{'}} + 50 y = 24 sin (10 t)

y = e^{- 6 t} (c_{1} cos (8 t) + c_{2} sin (8 t)) - \frac{2}{5} cos (10 t)

Solution steps

0.5 y^{''} (t) + 6 y^{'} (t) + 50 y = 24 sin (10 t)

Solve linear ODE:

y = e^{- 6 t} (c_{1} cos (8 t) + c_{2} sin (8 t)) - \frac{2}{5} cos (10 t)

y = e^{- 6 t} (c_{1} cos (8 t) + c_{2} sin (8 t)) - \frac{2}{5} cos (10 t)

y^{^{''}} + 2 y^{^{'}} + 2 y = 2

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

y^{^{''}} - 2 y^{^{'}} + 2 y = 1, y^{^{'}} 0 = 0

y^{^{''}} - 2 y^{^{'}} + y = \frac{7 e ^{t}}{1 + t ^{2}}

y^{^{''}} - 9 y = 9 e^{3 x} - 6 x^{4}

What is the solution for 0.5y^{''}+6y^'+50y=24sin(10t) ?
The solution for 0.5y^{''}+6y^'+50y=24sin(10t) is y=e^{-6t}(c_{1}cos(8t)+c_{2}sin(8t))-2/5 cos(10t)