Note that c2 really stands for c2/c1 in the problem definition

Clear[x];
DSolve[{x'[t]==((-1)/T)(x[t]-xa)-((k1^2)/c2)EPD (E^(((t-tf))/T)),x[t0]==x0},x[t],t]//Simplify

Expand[(((E^(-((t+tf)/T))) (-(E^((2 t)/T)) EPD (k1^2) T+(E^((2 t0)/T)) EPD (k1^2) T+2 c2 (E^((t0+tf)/T)) (x0-xa)+2 c2 (E^((t+tf)/T)) xa))/(2 c2))]

(E^(-((t+tf)/T)+((t0+tf)/T))) x0-(E^(-((t+tf)/T)+((t0+tf)/T))) xa//Simplify

-(((E^(((2 t)/T)-((t+tf)/T))) EPD (k1^2) T)/(2 c2))+(((E^(((2 t0)/T)-((t+tf)/T))) EPD (k1^2) T)/(2 c2))//Simplify

(E^(-((t+tf)/T))) ((E^((2 t)/T))+(E^((2 t0)/T)))//Simplify

Lets check this solution :

x[t_]:=xa+(x0-xa) (E^((-(t+t0))/T))-EPD (k1^2)(T/(2 c2))( (E^(((t-tf))/T))+(E^(((2 t0-t-tf))/T)));
D[x[t],t]+(1/T)(x[t]-xa)+((k1^2)/c2)EPD (E^(((t-tf))/T))//Simplify

Evaluate the above at tf and we get

x[tf]//Simplify

Solve[-(((1+(E^((2 (t0-tf))/T))) (x1f-xd) (k1^2) T)/(2 c2))+(E^(-((t0+tf)/T))) (x0-xa)+xa==x1f,x1f]//Simplify