Second Order Homogeneous Differential Equation.

[affirmative]

I would very much appreciate if anyone can help me with this problem; I have approached it from many different angles to no avail.

The position x(t) of a particle moving along the x-axis is governed by the differential
equation:

$$x'' + kx' + (n^2)x = 0$$ , and initially $$x(0) = a$$, $$x'(0) = u$$.

Show that:

$$\int_{0}^{Infinity} x^{2}dt = \frac{1}{2kn^2}((u + ka)^2 + n^2a^2)$$

Show that, as a function of k, this is a minimum when:

$$k^2 = n^2 + \frac{u^2}{a^2}$$

 

[gabbagabbahey]

You have a pretty straight forward constant coefficient 2nd order ODE. The first step is to solve it for x(t)...Have you at least been able to do that? What have you tried?

 

[affirmative]

$$x'' + kx' + (n^2)x = 0$$, let $$x = ce^{Yx}$$

So: $$x' = Yce^{Yx}$$
And: $$x'' = (Y^2)ce^{Yx}$$

So eventually: $$Y^2 + kY + n^2 = 0$$

I believe the next step would be to factorise this, and then depending on whether there are two roots or a repeated root it would be of the form $$y = Ae^{(Y1)x} + Be^{(Y2)x}$$ or $$y = (A + Bx)e^{Yx}$$ respectively. However, I am unsure what to do with the quadratic I obtained which contains constants k and n. Any ideas? Again, thanks for taking the time to help me.

Oh, and sorry it takes me so long to reply; I'm still getting the hang of this LaTeX thing...

 

[gabbagabbahey]

$$x'' + kx' + (n^2)x = 0$$, let $$x = ce^{Yx}$$

So: $$x' = Yce^{Yx}$$
And: $$x'' = (Y^2)ce^{Yx}$$

So eventually: $$Y^2 + kY + n^2 = 0$$

I believe the next step would be to factorise this, and then depending on whether there are two roots or a repeated root it would be of the form $$y = Ae^{(Y1)x} + Be^{(Y2)x}$$ or $$y = (A + Bx)e^{Yx}$$ respectively. However, I am unsure what to do with the quadratic I obtained which contains constants k and n. Any ideas? Again, thanks for taking the time to help me.

Oh, and sorry it takes me so long to reply; I'm still getting the hang of this LaTeX thing...

Just use the quadratic equation; you should get two distinct roots.

 

[affirmative]

The general solution I obtain:

$$y = Ae^{(1/2)(-b + \sqrt[2]{k^2-4n^2})x} + Be^{(1/2)(-b - \sqrt[2]{k^2-4n^2})x}$$

I was able to get this far when doing it by myself, it is here that I hit the wall. I fail to see what anything I have done so far has to do with proving the Integral at the beginning involving x^2.

 

[gabbagabbahey]

First, you should be obtaining x(t) not y(x); and second -b=-k

$$\Rightarrow x(t)=Ae^{\frac{-k + \sqrt{k^2-4n^2}}{2}t} + Be^{\frac{-k - \sqrt{k^2-4n^2}}{2}t}$$

Follow?

From here, you apply your initial conditions to determine A and B. Once you've done that, you can compute the integral $$\int_0^{\infty} x^2 dt$$ and show (after a fair bit of work) that you get what you're supposed to get.

 

[affirmative]

Maybe I just don't get differential equations - I don't know. All I manage to get for my constants A and B is the one in terms of the other and a or u.

x(0) yields: $$A+B=a$$
x'(0) yields: $$(1/2)[k(-A-B) + \sqrt[2]{k^2-4n^2}(A-B)] = u$$

As a result of this I can only get the constants in terms of two other constants; I cannot solve like I usually would if I had actual integers instead of letters.

 

[gabbagabbahey]

You'll need to solve for A and B in terms of 'u' and 'a' only...you have two equations, and two variables which you wish to solve for (A and B). So...?

 

[affirmative]

I'm at a total loss for what to do next. I think and think but nothing occurs to me. I don't just have two variables, no matter what I do I will always get A and B in terms of two other things; I can never solve for them since there are two letters a and u and not numbers.

 

[gabbagabbahey]

You should find that A=f(a,u) and B=g(a,u) where f and g are some functions. Your answers will be in terms of a and u.

For example, if I had the equations $$2A-B=u$$ and $$A+3B=7a-10u$$, I would solve the first one for B to get $B=2A-u$ and then substitute it into the second to get:

$$A+3(2A-u)=7a-10u \quad \implies A=a-u$$ and $$B=2A-u=2a-3u$$

Do you follow?

 

[affirmative]

Actually yes... Surprisingly. Just looking at it now I can understand what you meant when you said a fair amount of work! Thank you.