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I am trying to solve the set of differential equations

$$\dot{x} = \cos y\,,\quad\dot{y} = -\tanh x\sin y$$

for $x(t)$ and $y(t)$. One method I've encountered to decouple the equations involves turning this into a second-order differential equation via differentiation and substitution. But I've also tried dividing the equations to get a relation between $x$ and $y$ like this:

$$\frac{\dot{x}}{\dot{y}} = \frac{dx}{dt}\frac{dt}{dy} = \frac{dx}{dy} = -\frac{\cot y}{\tanh x}$$

since neither depends explicitly on $t$, which gives me a new relation:

$$A\cosh x = \sin y$$

where $A$ is a constant of integration. I then get

$$\ddot{x} \sim \sinh x\cosh x\,,\quad\ddot{y}\sim\sin y\cos y$$

where I've dropped leading constants. Is this a valid method? If I decouple in the normal way, where I take the derivative of $\dot{x}$ and substitute $\dot{y}$ into the first equation, I instead get the equation

$$\ddot{x} \sim \tanh x(1-\dot{x}^2)$$

Where am I going wrong? Will both methods give me the correct solution?

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    It looks OK to me, I don't see any mistake. Note that there can be different ways of writing an ODE for the same function. For example if $x(t) = \cos(t)$ then $x$ satisfy both $x'' + x = 0$ and $(x')^2 + x^2 = 1$. As for finding the solution: I would not be surprised if it turns out to be impossible to solve this in closed form. You might be able to write the solution on implicit form $g(x(t)) = t$ for some function $g(\cdot)$. – 2017-02-02
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    Using this method I get closed form solutions in terms of the Jacobi amplitude function. – 2017-02-02
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    Ok, nice. Though that is a highly non-elementrary function (basically defined as the inverse of some elliptic integral), but it's likely as closed form as you are going to get. – 2017-02-02

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The beginning of your calculus is correct.

If I made no mistake, the equation $\quad A\cosh(x)=\sin(y)\quad$ is false. It should be : $$A\cosh(x)=\frac{1}{\sin(y)}$$ Then, putting this result into the system of two ODEs leads to $x(t)$ and $y(t)$

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