#(dv)/(dt) + lambda v =e^(-alpha/3t)# ?

1 Answer
Mar 21, 2017

See below.

Explanation:

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)#