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It's been awhile since I took differential equations. Now I am using differential equations in another class. This is why you shouldn't sell back books from your major courses. :) How would I solve the following differential equation for x(t), y(t), z(t)?

$x'=(1-z^2)y$

$y'=(1-z^2)x$

$z'=0$

Also, the exercise I am looking at is on $S^2$, so $x^2+y^2+z^2=1$. I am not sure if this is relavant but I am telling you just in case.

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In this case, $z$ is constant. We have $x''=(1-z^2)y'=(1-z^2)^2 x$. This is a standard type of DE, with well-known solutions. The cases $z^2=1$ are particularly simple. For $z^2\ne 1$, we get linear combinations of exponentials.

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it's not a linear system, so a general method for this sort would not be the first thing you might learn about. let's see if we can be clever enough to solve.

solve the z'=0 first. you might end up with a linear system then. their are general methods for that, maybe something about diagonalizing the operator, but there may exist other tricks.

can you see a trick for solving something of the type x'=y, y'=x? try to find a differential equation with fewer dependent variables.

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First, $z=C_1$ is constant.

Then, divide the first equation by the second one, you get $$\frac{dx}{dy}=\frac{y}{x}$$ $$\int xdx=\int ydy$$ $$x^2+C_2=y^2$$ Substitute it back into any of the first two equations and solve it.

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