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Elementary differential equations

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data

    1. For the system of equations x'(t) = 4y and y'(t) = x, obtain the equation of the trajectory (path in the phase plane) that passes through (2, 0). For this trajectory, what is the equation of the slant asymptote that (x(t), y(t)) approaches as t goes to infinity?

    2. Verify that the given function is a solution to the given partial differential equation:

    [itex]\alpha^{2}u_{xx} = u_{t}[/itex];
    [itex]u_{1}(x, t) = e^{-\alpha^{2}t}sin x[/itex];
    [itex]u_{2}(x, t) = e^{-\alpha^{2}\lambda^{2}t}sin(\lambda x)[/itex]

    where [itex]\lambda[/itex] is a real constant. Nothing is said about [itex]\alpha[/itex], but I assume it is a constant?

    2. Relevant equations

    None

    3. The attempt at a solution

    1. I found dy/dx, which was separable, and integrated, and with the given initial condition, I ended up with [itex]2y^{2} = \frac{x^{2}}{2} - 2[/itex]

    Finding the slant asymptote as t goes to infinity...I'm not sure, since we don't have explicit formulas for x(t) and y(t). The given equations for x' and y' are not separable. I'm not sure where to go from here...any hints or help would be appreciated.

    2. I end up with:

    [itex]u_{1x} = e^{-\alpha^{2}t}cos x[/itex];
    [itex]u_{1xx} = -e^{-\alpha^{2}t}sin x[/itex];
    [itex]u_{1t} = -\alpha^{2}e^{-\alpha^{2}t}cos x[/itex];
    [itex]u_{1tt} = \alpha^{4}e^{-\alpha^{2}t}cos x[/itex];
    and then
    [itex]u_{2x}(x, t) = \lambda e^{-\alpha^{2}\lambda^{2}t}cos(\lambda x)[/itex]
    [itex]u_{2xx}(x, t) = -\lambda^{2} e^{-\alpha^{2}\lambda^{2}t}cos(\lambda x)[/itex]
    [itex]u_{2t}(x, t) = -\alpha^{2}\lambda^{2}e^{-\alpha^{2}\lambda^{2}t}sin(\lambda x)[/itex]
    [itex]u_{2t}(x, t) = \alpha^{4}\lambda^{4}e^{-\alpha^{2}\lambda^{2}t}sin(\lambda x)[/itex]

    and ALL I do is just use my given [itex]\alpha^{2}u_{xx} = u_{t}[/itex] and substitute each one into the other (staying with respective eqn 1 and eqn2 of course) and show that they're equal, right? Just wanted to confirm.

    Thanks in advance everyone!
     
  2. jcsd
  3. Jan 25, 2012 #2
    For 1) You should be able to solve the system from the start by taking another d/dt and setting it equal to the derivative in the other.

    y'' = x' (after taking d/dt of y')
    x' = 4y (you already have this)

    y'' = 4y (this is a solvable ODE which you can use, and then do similar for x or use your result for y to find x)

    Part 2 looks right to me.
     
  4. Jan 25, 2012 #3
    I did that and I get x = -1/(t+D) and y = -1/(t+C) (C, D are both constants).

    Substitute back into earlier known equations and we have x' = 2x^2 = 2/(t+D)^2 and y' = 2y^2 = 2/(t+C)^2. In both, as t goes to infinity, x' and y' both go to zero. Which means the asymptote is just y = x = 0, which doesn't pass through (2,0) in our initial condition.

    Where did I go wrong?
     
  5. Jan 25, 2012 #4

    vela

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    You didn't solve y''-4y=0 correctly. Your solution for y(t) doesn't satisfy the differential equation. You should get exponentials or, equivalently, a combination of sinh and cosh.

    Once you have x(t) and y(t) in explicit form, you'll see they tend to infinity as t does. To find the asymptotes, note that when x becomes large, you can neglect the constant on the righthand side of
    $$2y^2 = \frac{x^2}{2} - 2,$$so you have ##2y^2 \approx \frac{x^2}{2}##. Alternately, you can rewrite the equation as
    $$\frac{x^2}{4} - y^2 = 1$$ and identify it as that of a hyperbola, then use your knowledge of hyperbolas to find the asymptotes.

    The asymptotes don't satisfy the initial conditions. Only the solution (x(t), y(t)) needs to.
     
    Last edited: Jan 25, 2012
  6. Jan 25, 2012 #5
    Interestingly enough, I have never formally studied hyperbolas. In highschool, (11th and 12th grade particularly), it was taken out of the curriculum. In college, I've taken 2 years of calculus, a real analysis class, linear algebra, and now taking elementary DEs, and I have never, ever been required to know much, if anything about hyperbolas.

    Anyways, from what I know, the slope of the asymptote will be +/-1/4, correct? Or maybe i'm wrong...
     
  7. Jan 25, 2012 #6

    vela

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    Close. The slope is ±1/2. Do you see why?
     
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