What is the solution for y^{''}+4y=-8sin(2t) ?

The solution for y^{''}+4y=-8sin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+2tcos(2t)

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y^{''}+4y=-8sin(2t)

y^{^{''}} + 4 y = - 8 sin (2 t)

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

Solution steps

y^{''} (t) + 4 y = - 8 sin (2 t)

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2} x (cos (x) - 9 sin (x))

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} + y = 5 e^{t}

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

y^{^{''}} + 4 y = 3, 2 sin (4 t) y (0) = 1

y^{^{''}} - y^{^{'}} - 6 y = x e^{x}

What is the solution for y^{''}+4y=-8sin(2t) ?
The solution for y^{''}+4y=-8sin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+2tcos(2t)