What is the solution for y^{''}+4y=-20sin(2t) ?

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

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y^{''}+4y=-20sin(2t)

y^{^{''}} + 4 y = - 20 sin (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 5 t cos (2 t)

Solution steps

y^{''} (t) + 4 y = - 20 sin (2 t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 5 t cos (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + 5 t cos (2 t)

y^{^{''}} - 6 y^{^{'}} + 9 y = 2 e^{x}

x^{^{''}} + 2 x^{^{'}} + 5 x = 1, x^{^{'}} (0) = 0, x (0) = 0

y^{^{''}} + 4 y = 10 cos (2 x) + 15 sin (2 x)

y^{^{''}} - 5 y^{^{'}} = 3 cos (x)

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2} x (cos (x) - 8 sin (x))

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