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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + 7 y = 5 t^{2} - 5, y (0) = 0, y^{^{'}} (0) = - 2

y^{^{'''}} + y^{^{'}} = sec^{2} (x)

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 9 + 2 sin (x)

y^{^{''}} - y = e^{x} sin (x)

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