Hello !

**Answer.**

#y(x) = A x^(-1) + B cos(sqrt(3)/2 x) + C sin(sqrt(3)/2 x)+ 2 - (cos(x)+sin(x))/2#

where #A,B,C# are real constants.

- First, solve the homogeneous equation : #y'''+y=0#.

The characteristic equation is #x^3+1=0#, the complex roots are #-1,-j,-j^2#, where #j=e^{i (2 pi)/3} = -1/2+isqrt(3)/2# (famous number !)

So, the solutions of #y'''+y=0# are the functions :

#x \mapsto lambda x^{-1} + mu x^{-j} + nu x^{-j^2}#

where #lambda, mu, nu# are arbitrary complex constants.

Maybe you wish real solutions. It's not a problem. Take the real part.

1) #x^{-j} = e^{-j ln(x)} = e^{(1/2 - i sqrt(3)/2) ln(x)} = sqrt(x) e^{-i sqrt(3)/2 ln(x)}#.

2) Same thing with #x^{-j^2}#, with #-j^2 = 1/2 + i sqrt(3)/2# :

#x^{-j^2} = sqrt(x)e^{i sqrt(3)/2 ln(x)}#.

The real solutions of #y'''+y=0# are :

#x \mapsto A x^(-1) + B cos(sqrt(3)/2 x) + C sin(sqrt(3)/2 x)#

where #A,B,C# are real constants.

- Second, find a particular solution of #y'''+y = 2-sin(x)#. To do that,

1) find a particular solution (so called #f_1#) of #y'''+y=2#

2) find a particular solution (so called #f_2#) of #y'''+y=sin(x)#

A particular solution of #y'''+y = 2-sin(x)# will be #f_1-f_2#.

1) To find #f_1#, it's really easy : take #f_1(x) = 2# (constant function).

2) To find #f_2# it's more clever. Remember that

#(d^3)/(d x^3) cos(x) = sin(x)# and #(d^3)/(d x^3) sin(x) = -cos(x)#,

so you have

#(d^3)/(d x^3) (cos(x)+sin(x)) = 2 sin(x)#, therefor :

#(d^3)/(d x^3) ((cos(x)+sin(x))/2) = sin(x)#.

So, you can take #f_2(x) = (cos(x)+sin(x))/2#.

Finally, a particular solution of #y'''+y = 2-sin(x)# is

#x \mapsto 2 - (cos(x)+sin(x))/2#.

**Conclusion.** All the (real) solutions of #y'''+y = 2-sin(x)# are

#x \mapsto A x^(-1) + B cos(sqrt(3)/2 x) + C sin(sqrt(3)/2 x)+ 2 - (cos(x)+sin(x))/2#

where #A,B,C# are real constants.