What is the solution for y^{''}+4y=-16sin(2t) ?

The solution for y^{''}+4y=-16sin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+4tcos(2t)

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y^{''}+4y=-16sin(2t)

y^{^{''}} + 4 y = - 16 sin (2 t)

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

Solution steps

y^{''} (t) + 4 y = - 16 sin (2 t)

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 13 y = 52, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 e^{2 x}

y^{^{''}} - y^{^{'}} + y = 5 t^{2}

x^{^{''}} + 2 x^{^{'}} + 5 x = 1

q^{^{''}} + 1100 q^{^{'}} + 100000 q = 16

What is the solution for y^{''}+4y=-16sin(2t) ?
The solution for y^{''}+4y=-16sin(2t) is y=c_{1}cos(2t)+c_{2}sin(2t)+4tcos(2t)