Is this differential equation correctly calculated as having as its solution this parametric family of curves? 
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:
$$x'(t)=\frac{x(t)\cdot t}{1+t\cdot\tan{t}}$$
$$y'(t)=\frac{y(t)(2\cos{t}-t\dot\sin{t})}{\sin{t}+t\cdot\cos{t}}$$
 A: 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?
