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

The solution for y^{'''}+y^'=-2+2sin(x) is y=c_{1}+c_{2}cos(x)+c_{3}sin(x)-2x-xsin(x)

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

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

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) - 2 x - x sin (x)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) - 2 x - x sin (x)

y = c_{1} + c_{2} cos (x) + c_{3} sin (x) - 2 x - x sin (x)

y^{^{''}} - 2 y^{^{'}} + 2 = 0

y^{^{''}} - 2 y^{^{'}} + 82 y = e^{x} cos (9 x)

y^{^{''}} - 2 y^{^{'}} - 3 y = 5 e^{4 x}, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 6

y^{^{''}} - y = 2, y (0) = 2, y^{^{'}} (0) = 0

What is the solution for y^{'''}+y^'=-2+2sin(x) ?
The solution for y^{'''}+y^'=-2+2sin(x) is y=c_{1}+c_{2}cos(x)+c_{3}sin(x)-2x-xsin(x)