Answer:
Step-by-step explanation:
Characteristic equation:
[tex]\lambda^2-2\lambda + 5 =0[/tex]
[tex]\lambda = 1\pm2i[/tex]
The solution of the ODE is
[tex]y = e^x(C_1 cos(2x) + C_2 sin(2x) )[/tex]
Now, very find by taking the first and the second derivative of y.
[tex]y' = e^x(-2 C_1 sin (2x) +2C_2 cos (2x)) +e^x (C_1 cos (2x) + C_2sin(2x))[/tex]
[tex]=-2e^xC_1 sin(2x) +2e^xC_2 cos (2x)+C_1e^xcos(2x) + C_2e^x sin(2x)[/tex]
[tex]y" = -2e^x C_1 sin(2x) -4e^x C_1cos(2x)+2e^xC_2cos(2x)-4e^xC_2sin(2x) +C_1e^x cos(2x) -2C_1e^xsin(2x)+C_2e^x sin(2x) +2C_2e^x cos(2x)[/tex]
[tex]=-4C_1e^xsin(2x) -3C_1e^x cos(2x) +4C_2e^xcos(2x) -3C_2e^xsin(2x)[/tex]
Now, put all in y" -2y'+5y and consider if it = 0 or not.