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.
