Question

Using method of variation find solution of y"+2iy-y=e^x-2e^ix

Answer 1

I guess you mean: y"+2iy'-y=e^x-2ie^x.

First solve y"+2iy'-y=0 to get the characteristic equation, y_c.

We have (r+i)²=0, because y"+2iy'+i²y=y"+2iy'-y=0.

y_c=Ae^-ix+Bxe^-ix. Note that e^-ix=cos(x)-isin(x).

The particular solution, y_p=u₁e^-ix+u₂xe^-ix, where u₁ and u₂ are functions of x. We find them using Wronskian determinants w, w_1, w₂, and Cramer's Rule for solving simultaneous equations.

w=

|   e^-ix     xe^-ix      |

| -ie^-ix -ixe^-ix+e^-ix | = e^-2ix. Note that e^-2ix=cos(2x)-isin(2x).

w₁=

|      0         xe^-ix      |

| e^x-2ie^x -ixe^-ix+e^-ix | = -xe^x-ix+2ixe^x-ix=xe^x-ix(-1+2i).

w₂=

|   e^-ix      0     |

| -ie^-ix e^x-2ie^x | = e^x-ix-2ie^x-ix=e^x-ix(1-2i). Note that e^x-ix=e^x(cos(x)-isin(x)).

u₁=w₁/w=xe^x+ix(-1+2i); u₂=w₂/w=e^x+ix(1-2i).

y_p=xe^x+ix(-1+2i)e^-ix+xe^x+ix(1-2i)e^-ix=xe^x(-1+2i)+xe^x(1-2i)=0.

Therefore, y=y_c=Ae^-ix+Bxe^-ix=(A+Bx)(cos(x)-isin(x)) or (A+Bx)e^-ix.