Is this differential equation correctly calculated as having as its solution this parametric family of curves?

Question

Consider the following family of curves is parametric form:

$$x(t) = a(\cos{t} + t\sin{t})$$ $$y(t) = a(\sin{t} + t\cos{t})$$

Where $a\in\mathbb{R^+}$ is a constant, and $t\geq0$. Find the differential equation which verifies such family formulae.

So, what I did, was differentiate both equations in terms of t, then isolate the constant a from the differentiated formula, and plug it back into the original equations to get rid of the constant, hence ending up with the original formulae only in terms of $x, y, x', y'$ and $t$.

First of all, I'd like to know whether this process is correct, and if it does indeed yield all solutions (if it's even correct). The actual solutions I got are:

$$x'(t)=\frac{x(t)\cdot t}{1+t\cdot\tan{t}}$$ $$y'(t)=\frac{y(t)(2\cos{t}-t\dot\sin{t})}{\sin{t}+t\cdot\cos{t}}$$

Answer 1

The differential equation for $y$ is wrong. $y = a (\sin(t)-t\cos(t))$ does not satisfy it.

EDIT: with the changed formula for $y$ it's correct.

Another problem is that there's no reason the $a$ for a solution of the first differential equation should be the same as the $a$ for a solution of the second differential equation.

EDIT: A system of two (non-autonomous) differential equations for $x'$ and $y'$ will have a two-parameter family of solutions. Here you want a one-parameter family. So this won't work. What you would need would be autonomous differential equations. You'd have to eliminate both $a$ and $t$ to get $x'$ and $y'$ as functions of $x$ and $y$. I don't think that can be done in closed form.

In principle you can (locally, except at singularities) write $$ \dfrac{dy}{dx} = \dfrac{y'}{x'} = f(y/x) $$ for a function $f$ defined implicitly, i.e. $$ \dfrac{2 \cos(t) - t \sin(t)}{t \cos(t)} = f \left( \dfrac{\cos(t)+t\sin(t)}{\sin(t)+t\cos(t)} \right) $$

I suspect that's the kind of solution that's expected. But are you sure you have the question right?