What is the solution for y^{''}-2y^'+5y=20sin(t) ?

The solution for y^{''}-2y^'+5y=20sin(t) is y=e^t(c_{1}cos(2t)+c_{2}sin(2t))+4sin(t)+2cos(t)

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y^{''}-2y^'+5y=20sin(t)

y^{^{''}} - 2 y^{^{'}} + 5 y = 20 sin (t)

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 20 sin (t)

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 6 x + 4

y^{^{''}} + 4 y = 6 sin (2 t), y (0) = 2, y^{^{'}} (0) = 1

10 y^{^{''}} + 1000 y = 160 cos (8 t)

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

y^{^{''}} + y^{^{'}} - 2 y = 13 + e^{x}

What is the solution for y^{''}-2y^'+5y=20sin(t) ?
The solution for y^{''}-2y^'+5y=20sin(t) is y=e^t(c_{1}cos(2t)+c_{2}sin(2t))+4sin(t)+2cos(t)