The differential equation describing the truck movement is a linear non homogeneous differential equation
#(dv)/(dt) + lambda v =e^(-alpha/3t)#
The solution for those type of equations can be obtained as the sum of two solutions. The solution to the homogeneous equation
#(dv_h)/(dt)+lambda v_h = 0#
plus the particular solution
#(dv_p)/(dt) + lambda v_p =e^(-alpha/3t)#
After that, #v = v_h + v_p#
Obtaining #v_h# is quite easy
We propose #v_h = C e^(xi t)# then substituting into the homogeneous
#lambda C e^(xi t)+xi C e^(xi t)=C(lambda+ xi)e^(xi t)=0#
This condition is satisfied for #lambda + xi = 0# or
#xi = -lambda#
and #v_h = C e^(-lambda t)#
The particular is obtained supposing that #C = C(t)# and introducing into the complete equation
#d/(dt)(C(t)e^(-lambda t))+lambda C(t)e^(-lambda t)=e^(-alpha/3t)#
so we obtain
#(dC)/(dt)e^(-lambda t)=e^(-alpha/3 t)#
then
#(dC)/(dt) = e^((lambda-alpha/3)t)# and integrating
#C(t) = e^((lambda-alpha/3)t)/(lambda-alpha/3)#
and finally
#v = C_0 e^(-lambda t)+e^((lambda-alpha/3)t)/(lambda-alpha/3)e^(-lambda t) = C_0 e^(-lambda t)+e^(-alpha/3t)/(lambda-alpha/3)#
