Because Characteristic equation of homogenius equation: r2 - 1 = 0, r = ± 1.
Then general solution of homogenius solution is yh = C1ex :+ C2e-x.
Particular solution of given equation is yp = (ax + b)e2x, where a and b is unknown.
Now yp' = ae2x + 2(ax + b)e2x = (2ax + 2b + a)e2x; yp'' = 2ae2x + 2(2ax + 2b + a)e2x = (4ax + 4b + 4a)e2x.
Substitution in given equation after reducing by e2x we get: 4ax + 4b + 4a - ax - b = x or 3ax + 3b + 4a = x.
So, a = 1/3, 3b + 4/3 = 0, b = - 4/9and because yp = (1/3x - 4/9)e2x and general solution of given ODE is:
y = C1ex + C2e-x + (1/3x - 4/9)e2x