The corresponding homogeneous ODE has characteristic equation [tex]r^2-r+1=0[/tex] with roots at [tex]r=\dfrac{1\pm\sqrt3}2[/tex], thus admitting the characteristic solution
[tex]y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x[/tex]
For the particular solution, assume one of the form
[tex]y_p=a\sin x+b\cos x[/tex]
[tex]{y_p}'=a\cos x-b\sin x[/tex]
[tex]{y_p}''=-a\sin x-b\cos x[/tex]
Substituting into the ODE gives
[tex](-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x[/tex]
[tex]-b\cos x+a\sin x=\sin x[/tex]
[tex]\implies a=1,b=0[/tex]
Then the general solution to this ODE is
[tex]\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}[/tex]
[tex]\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2[/tex]
[tex]\implies y_c=C_1e^x+C_2e^{2x}[/tex]
Assume a solution of the form
[tex]y_p=e^x(a\sin x+b\cos x)[/tex]
[tex]{y_p}'=e^x((a+b)\cos x+(a-b)\sin x)[/tex]
[tex]{y_p}''=2e^x(a\cos x-b\sin x)[/tex]
Substituting into the ODE gives
[tex]2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x[/tex]
[tex]-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x[/tex]
[tex]\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12[/tex]
so the solution is
[tex]\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}[/tex]
[tex]r^2+1=0\implies r=\pm i[/tex]
[tex]\implies y_c=C_1\cos x+C_2\sin x[/tex]
Assume a solution of the form
[tex]y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)[/tex]
[tex]{y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)[/tex]
Substituting into the ODE gives
[tex](-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)[/tex]
[tex]-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)[/tex]
[tex]\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49[/tex]
so the solution is
[tex]\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}[/tex]
