Q11E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Example 1, we solved for the velocity of the object as a function of time (equation (5)). In some cases, it is useful to have an expression, independent of t, that relates v and x. Find this relation for the motion in Example 1. [Hint: Letting $\mathbf{v}\left(t\right)\mathbf{=}\mathbf{V}\left(x\left(t\right)\right)$, then $\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{t}}\mathbf{=}\left(\frac{dv}{dx}\right)\mathbf{V}$]

Step-by-Step Solution

Answer

The relation for the motion is $e^{bV}{\left|mg-bV\right|}^{mg}=e^{b{v}_0}{\left|mg-b{v}_0\right|}^{mg}{e}^{\frac{-b^{2}x}{m}}$.

Step 1: Important hints.

To solve this question, use chain rule, and integration rules.

Step 2: Find the equation of motion for this relation

Here given is $v\left(t\right)=V\left(x\left(t\right)\right)$

Apply chain rule for the solution $\frac{dv}{dt}=\frac{dv}{dx}.\frac{dx}{dt}$

$v\frac{dv}{dx}=V\frac{dx}{dt}$

Now

 $$\begin{gathered} mV\frac{dv}{dt}=mg-bv \\ \frac{ mVdV}{mg-bV}=dx \left(\mathbf{Variableseparating}\right) \\ \frac{mVdV}{mg-bV}=dx \\ m\int \frac{Vdv}{mg-bV}=x+C \end{gathered}$$

Step 2: Finding the value of ∫ Vdv mg - bV

Now, 

$m\int \frac{Vdv}{mg-bV}=\left|mg-bV=tdV=-\frac{dt}{b}\right|\left(V=\frac{mg-t}{b}\right)$

$$\begin{gathered} \frac{1}{b^2}\int \frac{t-mg}{t}dt \\ \frac{1}{b^2}\int dt-mg\int \frac{dt}{t} \\ \frac{1}{b^2}(t-mg In|t\left|\right)+c_1 \\ \frac{1}{b^2}(mg-bV-mg In|mg-bV\left|\right)+c_1 \\ \frac{mg}{b^2}-\frac{V}{b}-\frac{mg\ln \left|mg-bV\right|}{b^2}+c_1 \end{gathered}$$

Now using then $C-\frac{mg}{b^2}=A$then

$$\begin{gathered} -\frac{V}{b}-\frac{gm \ln \left|mg-bV\right|}{b^2}+D=\frac{x}{m}+A \\ -bV-gm \ln \left|mg-bV\right|=\frac{b^{2}x}{m}+B \\ gm\ln \left|mg-bV\right|=-\frac{b^{2}x}{m}-bV-B \\ \ln {\left|mg-bV\right|}^{gm}=-\frac{b^{2}x}{m}-bV-B \\ {\left|mg-bV\right|}^{gm}=P{e}^{\frac{-b^{2}x}{m}}{e}^{-bV} \\ {e}^{bV}{\left|mg-bV\right|}^{gm}=P{e}^{\frac{-b^{2}x}{m}} \end{gathered}$$

When then $V\left(0\right)=v_0$ then $e^{b{v}_0}{\left|mg-b{v}_0\right|}^{mg}=P{e}^0$

Therefore, ${\mathbf{e}}^{\mathbf{bV}}{\left|\mathbf{mg}\mathbf{-}\mathbf{bV}\right|}^{\mathbf{mg}}\mathbf{=}{\mathbf{e}}^{{\mathbf{bv}}_{\mathbf{0}}}{\left|\mathbf{mg}\mathbf{-}{\mathbf{bv}}_{\mathbf{0}}\right|}^{\mathbf{mg}}{\mathbf{e}}^{\frac{\mathbf{-}{\mathbf{b}}^{\mathbf{2}}\mathbf{x}}{\mathbf{m}}}$