What is the solution for 0.5y^{''}+6y^'+50y=240cos(10t) ?

The solution for 0.5y^{''}+6y^'+50y=240cos(10t) is y=e^{-6t}(c_{1}cos(8t)+c_{2}sin(8t))+4sin(10t)

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0.5y^{''}+6y^'+50y=240cos(10t)

0.5 y^{^{''}} + 6 y^{^{'}} + 50 y = 240 cos (10 t)

y = e^{- 6 t} (c_{1} cos (8 t) + c_{2} sin (8 t)) + 4 sin (10 t)

Solution steps

0.5 y^{''} (t) + 6 y^{'} (t) + 50 y = 240 cos (10 t)

Solve linear ODE:

y = e^{- 6 t} (c_{1} cos (8 t) + c_{2} sin (8 t)) + 4 sin (10 t)

y = e^{- 6 t} (c_{1} cos (8 t) + c_{2} sin (8 t)) + 4 sin (10 t)

y^{^{''}} + y = 4 \cdot cos (x)

y^{^{''''}} - y = 8 t + cos (t)

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

y^{^{''}} - 4 y^{^{'}} + 5 y = e^{- 3 t}, y (0) = 1, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} + 100 y = csc (10 x)

What is the solution for 0.5y^{''}+6y^'+50y=240cos(10t) ?
The solution for 0.5y^{''}+6y^'+50y=240cos(10t) is y=e^{-6t}(c_{1}cos(8t)+c_{2}sin(8t))+4sin(10t)