The characteristic equation for this ODE is
[tex]r^2+2r+5=0[/tex]
which has roots at [tex]r=-1\pm2i[/tex], so the general solution is
[tex]y_c=(C_1\cos2x+C_2\sin2x)e^{-x}[/tex]
Given [tex]y(0)=2[/tex], we have
[tex]2=(C_1\cos0+C_2\sin0)e^{-0}\implies C_1=2[/tex]
and [tex]y'(0)=2[/tex], we have (upon differentiating [tex]y_c[/tex])
[tex]{y_c}'=((2C_2-C_1)\cos2x-(2C_1+C_2)\sin2x)e^{-x}[/tex]
[tex]2=((2C_2-2)\cos0-(4+C_2)\sin0)e^{-0}[/tex]
[tex]2=2C_2-2\implies C_2=2[/tex]
So the particular solution is
[tex]y_c=(2\cos2x+2\sin2x)e^{-x}[/tex]
