There are two DEs because the real and imaginary parts must each equate to zero.
(1) y(3)+y(1)=0 and (2) y(2)+y=0.
Let z=y(1), then (1) becomes z(2)+z=0.
If z=Asin(x)+Bcos(x), then z(1)=Acos(x)-Bsin(x) and z(2)=-Asin(x)-Bcos(x)=-z, therefore z(2)+z=0.
y(1)=Asin(x)+Bcos(x), y=-Acos(x)+Bsin(x)+C (A, B, C are constants).
(2) y=Dsin(x)+Ecos(x), where D and E are constants.
y=-Acos(x)+Bsin(x)+C+i(Dsin(x)+Ecos(x)).
-iy=-i(-Acos(x)+Bsin(x)+C)+(Dsin(x)+Ecos(x)),
y(1)=Asin(x)+Bcos(x)+i(Dcos(x)-Esin(x)),
-iy(2)=-i(Acos(x)-Bsin(x))+(-Dsin(x)-Ecos(x)),
y(3)=-Asin(x)-Bcos(x)+i(-Dcos(x)+Esin(x)).
So y(3)-iy(2)+y(1)-iy=-iC=0, so C=0.
And y=-Acos(x)+Bsin(x)+i(Dsin(x)+Ecos(x)).
The real part is y=-Acos(x)+Bsin(x).