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:

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


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?

  • $\begingroup$ Oops, regarding the first one, that's totally my fault, I made a mistake when copying it here, it should be a plus sign. Regarding the second one, if this argument isn't solid, then I have no clue how to proceed, any hints? $\endgroup$ – F.Webber Feb 17 '15 at 22:17
  • $\begingroup$ I just read your edited answer, it surprises me that the problem doesn't seem to have solution. Maybe I'm doing something wrong with how I'm considering the problem? The formulation of the problem asks for "the differential equation", in singular, while I got that system of 2 equations. Do you have any clue how to see this problem from a different point of view? Maybe by expressing the family with a single implicit equation (just like a circle or an ellipse, or any other $\mathbb{R^2}$ for that matter), we could get a single differential equation? $\endgroup$ – F.Webber Feb 17 '15 at 22:54

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