What is the solution for y^{''}+4y=4cos(2x) ?

The solution for y^{''}+4y=4cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+xsin(2x)

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = e^{4 t}

y^{^{''}} - y^{^{'}} - 12 y = e^{2 x} + e^{x} - 2

y^{^{'''}} + 64 y^{^{'}} = sec (8 x)

y^{^{''}} + 9 y = 18 x^{2} + 9 x - 6

y^{^{''}} - 4 y^{^{'}} + 4 y = 4 t^{2}

What is the solution for y^{''}+4y=4cos(2x) ?
The solution for y^{''}+4y=4cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)+xsin(2x)