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

Mar 21, 2017

See below.

Explanation:

The differential equation describing the truck movement is a linear non homogeneous differential equation

$\frac{\mathrm{dv}}{\mathrm{dt}} + \lambda v = {e}^{- \frac{\alpha}{3} t}$

The solution for those type of equations can be obtained as the sum of two solutions. The solution to the homogeneous equation

$\frac{{\mathrm{dv}}_{h}}{\mathrm{dt}} + \lambda {v}_{h} = 0$

plus the particular solution

$\frac{{\mathrm{dv}}_{p}}{\mathrm{dt}} + \lambda {v}_{p} = {e}^{- \frac{\alpha}{3} t}$

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

$\lambda C {e}^{\xi t} + \xi C {e}^{\xi t} = C \left(\lambda + \xi\right) {e}^{\xi t} = 0$

This condition is satisfied for $\lambda + \xi = 0$ or

$\xi = - \lambda$

and ${v}_{h} = C {e}^{- \lambda t}$

The particular is obtained supposing that $C = C \left(t\right)$ and introducing into the complete equation

$\frac{d}{\mathrm{dt}} \left(C \left(t\right) {e}^{- \lambda t}\right) + \lambda C \left(t\right) {e}^{- \lambda t} = {e}^{- \frac{\alpha}{3} t}$

so we obtain

$\frac{\mathrm{dC}}{\mathrm{dt}} {e}^{- \lambda t} = {e}^{- \frac{\alpha}{3} t}$

then

$\frac{\mathrm{dC}}{\mathrm{dt}} = {e}^{\left(\lambda - \frac{\alpha}{3}\right) t}$ and integrating

$C \left(t\right) = {e}^{\left(\lambda - \frac{\alpha}{3}\right) t} / \left(\lambda - \frac{\alpha}{3}\right)$

and finally

$v = {C}_{0} {e}^{- \lambda t} + {e}^{\left(\lambda - \frac{\alpha}{3}\right) t} / \left(\lambda - \frac{\alpha}{3}\right) {e}^{- \lambda t} = {C}_{0} {e}^{- \lambda t} + {e}^{- \frac{\alpha}{3} t} / \left(\lambda - \frac{\alpha}{3}\right)$