compute all real solutions of differential equation y(3)-iy(2)+y(1)-iy=0

Question

how to procede to get the real solutions, i get the roots of the characteristic equation but cant proceed from there to get real solutions.

Answer 1

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).