Question

The motion of a set of particles moving along the x-axis is governed by the differential equation
dx/dt=t^3-x^3
where x(t) denotes the position at time t of the particle.

a. If a particle is located at x=1 when t=2, what is its velocity at this time?

b. Show that the acceleration of a particle is given by

d^2x/dt^2=3t^2-3t^3x^2+3x^5

c. If a particle is located at x=2 when t=2.5, can it reach the location x=1 at any later time?
(hint: t^3-X^3=(t-x)(t^2+xt+x^2)

Answer 1

Part (a) is fairly straightforward. Any idea of what to do here, to determine the velocity of the particle when x=1 and t=2?

Answer 2

I understand a. you just plug the values of x and t into the differential equation to get velocity: 2^3-1^3=7

Part b is the derivative of velocity differentiated implicitly with respect to t

But I am lost on part c??

Answer 3

I'll reply by posing a question for you:
Is it possible to find x as a function of t through integration?

If so, you could let x=1 and try solving for t. If there is a solution (or are solutions) to that, then you could say whether or not the particle ever returns to x=1.

Answer 4

b is

\[3t ^{2}-2x ^{2}(\frac{ dx }{ dt })\]

Answer 5

If you start with the velocity and want a formula for distance, would you differentiate or integrate?

Note: in Part c, you're given a hint: It says that when t=2.5, x=2.

Answer 6

Is this a differential equations course, or just an advanced calculus course?

Answer 7

I would integrate to get a formula for distance

Answer 8

it is the beginning of a differential equations course

Answer 9

i have to go to work right now but will be back later to try and solve this, thanks so far!!

Answer 10

When I integrated both sides I got:
\[\frac{ dx }{dt }=t ^{3}-x ^{3}\]
\[x ^{3}dx=t ^{3}dt\]
\[\int\limits x ^{3}dx=\int\limits t ^{3}dt\]
\[\frac{ 1 }{ 4}x ^{4}=\frac{ 1 }{ 4 }t ^{4}+c\]

Plugging in x=2 and t=5/2 I get c=-369/64

So if t is greater than 2.5 x can never be equal to 1

Answer 11

First: dx/dt=t^3-x^3 = (t-x)(t^2+xt+x^2) (This hint is in the problem statement).

Second: I see you've attempted to integrate dx/dt=t^3-x^3 as though it were a separable differential equation. I don't think it is one.

If:\[\frac{ dx }{ dt }=t ^{3}-x ^{3}\] we could multiply both sides by dt to obtain:\[dx=t ^{3}dt - x^3dt\]
but this does not convert to x^3 dx = t^3 dt, does it?

Answer 12

I'm feeling conflicted here. On one had you say you're just starting a diff'l. eq'ns. course, but on the other hand, more sophisticated methods than ours are needed to find a solution to this d. e. in two variables.

So I'm wondering whether there's a simpler, graphical approach, where you don't try to integrate dx/dt, but rather calculate the slope for a number of combos of x and t (such as x=2 and t=2.5). Have you done this before? I would think so, if you're beginning this d. e. course.

If you go back to the equation for dx/dt, and let x=2 and t=2.5, you'd get

dx/dt = (2.5)^3 - (2)^3, or 6.25 - 8, or -1.75. That's the same as -7/4. You could actually graph (2,2.5) and then draw in a short, downward sloping line to represent the slope of -7/4 at that point. If you were to do this again and again, getting closer to x = 1, and then sketch a rough graph that is always parallel to your little slope lines, you'd be able to see from the graph whether the graph would ever approach x=1 again.

Hope this helps. I've been on my computer for hours and am no longer thinking very clearly.

Answer 13

This problem is from the 3rd section of chapter 1 in differential equations. The section covers direction fields.

Answer 14

I'm pretty confused on part c of this problem.

When I plug in the values t=2.5 and x=2 I get dx/dt=(2.5)^3-2^3=7.625