Q8E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

The motion of a set of particles moving along the x‑axis is governed by the differential equation $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{,}$  where  $\mathbf{x}\left(\mathbf{t}\right)$ denotes the position at time t of the particle.

⦁    If a particle is located at  $\mathit{x}\mathbf{=}\mathbf{1}$ when  $\mathit{t}\mathbf{=}\mathbf{1}$ , what is its velocity at this time?

⦁    Show that the acceleration of a particle is given by  $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{5}}\mathbf{.}$

⦁    If a particle is located at  $\mathit{x}\mathbf{=}\mathbf{2}$ when $\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}$, can it reach the location  $\mathit{x}\mathbf{=}\mathbf{1}$ at any later time?

[Hint: ${\mathbf{t}}^{\mathbf{3}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{=}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{(}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{xt}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{)}\mathbf{.}$ ]

Step-by-Step Solution

Answer

⦁    7

⦁  $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{5}}$

⦁    No

1(a): Finding the velocity by putting x = 1 and t = 2 in dx dt

Given,  $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{,}$

Substituting  $x=1$ and $t=2$  , one gets,

 $$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}{\mathbf{2}}^{\mathbf{3}}\mathbf{-}{\mathbf{1}}^{\mathbf{3}} \\ \mathbf{=}\mathbf{8}\mathbf{-}\mathbf{1} \\ \mathbf{=}\mathbf{7} \end{gathered}$$

i.e., Velocity at time $t=2$  and position  $x=1$ is 7.

Hence, the velocity is 7.

2(b): Differentiating dx dt with respect to t.

 $\frac{\mathbf{dx}}{\mathbf{dt}}$ represents velocity while  $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}$ gives the acceleration of the particle.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{,} \\ \frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{.}\frac{\mathbf{dx}}{\mathbf{dt}} \\ \mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{(}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}\mathbf{)} \\ \mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{5}} \end{gathered}$$

Hence, it is shown that the acceleration of a particle is given by  $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{5}}\mathbf{.}$

3(c): Analyzing the position of particle at different time

From the graph, it is visible that the particle is stabilized about the velocity  $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{1}$

Now, if  $x=1$ and $t=2.5$, then  $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{>}\mathbf{0}$, i.e., the particle moves away from $x=1$ .

Thus, the position of the particle keeps on increasing and can reach x = 1 only once.