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stripes
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Homework Statement
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?
Homework Equations
None
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!