Why does solving the differential equation for circular motion lead to an illogical result? In uniform circular motion, acceleration is expressed by the equation
$$a = \frac{v^2}{r}. $$
But this is a differential equation and solving it gets the result $$v = -\frac{r}{c+t}.$$
This doesn’t makes any sense. Should velocity be constant? Or at least something trigonometric?
 A: The acceleration points towards the center of rotation, the acceleration is missing a $-\hat{r}$
The equations of motion for circular motion of radius $R$ at a constant frequency are
$$\boldsymbol{R}(t)= R \cos(\omega t) \hat i + R \sin(\omega t) \hat j$$
Velocity is changing, otherwise acceleration would be zero. Speed however, is constant.
Differentiate this and you obtain acceleration,
$$\boldsymbol{a} = \frac{d\boldsymbol{R}}{dt}= -\omega R \sin(\omega t) \hat{i} + \omega R \cos(\omega t) \hat{j} $$
$$\| \boldsymbol{a} \| = \sqrt{\omega^2 R^2 \sin^2(\omega t) + \omega^2 R^2 \cos^2(\omega t)} = \omega R$$
A: That equation is misleading.
You wrongly assume that $a$ is $dv/dt$ here, which it isn't.
Let us first rewrite the centripetal acceleration equation properly:
$$|\vec {\mathbf a}_N| = \dfrac {\left|\vec {\mathbf v}\right|^2} {R} \Longleftrightarrow a_N = \dfrac {v^2} R$$ where $a_N$ is the normal acceleration.  The normal acceleration can be expressed as $\vec{\mathbf a} - \mathbf{\vec a}_T$ where $a_T$ is the tangential compenent of acceleration $a$.  For uniform circular motion, $a_T=0$ so what we're left with is $$\left|\dfrac{d\mathbf{\vec v}}{dt}\right| = \dfrac{|\mathbf{\vec v}|^2}{R} \Longleftrightarrow a=\dfrac{v^2}R$$
Note that $a$ here is $a=\sqrt{a_x^2 +a_y^2}$ in 2-D. Since $\mathbf{\vec v} = \left( v_x, v_y\right)$, we can say $$a=\sqrt{\left(\dfrac{dv_x}{dt}\right)^2+\left(\dfrac{dv_y}{dt}\right)^2}.$$
Putting this into our centripital acceleration equation, we can get,
$$\sqrt{\left(\dfrac{dv_x}{dt}\right)^2+\left(\dfrac{dv_y}{dt}\right)^2}=\dfrac{v_x^2 + v_y^2 }R \tag{*}$$
This is the proper differential equation we are solving.
Note that since $a_T=0$, we have that $\dfrac d{dt} \left| \mathbf{\vec v}\right| = \dfrac d{dt} \sqrt{v_x^2 + v_y^2}=0$, so $v_x a_x + v_y a_y = \mathbf{\vec v}\cdot\mathbf{\vec a}=0$, which serves as our second equation to solve ($\star$).
Now, since you know that the solution to uniform circular motion is $\mathbf{\vec r} = \left( R\cos(\omega t), R\sin (\omega t\right))$, you can go ahead and verify that it indeed satisfies $\star$.
A: The equation that you've given is just a simplified version of the real thing that lacks all direction information. The real thing is this:
$$\vec{a} = -\frac{\vec{v}^2}{\vec{r}^2}\cdot\vec{r}$$
If you take the magnitude of the vector on both sides of this, the equation simplifies as
$$|\vec{a}| = \left| -\frac{\vec{v}^2}{\vec{r}^2}\cdot\vec{r} \right|$$
$$\Leftrightarrow |\vec{a}| = \frac{|\vec{v}|^2}{|\vec{r}|^2}\cdot|\vec{r}|$$
$$\Leftrightarrow |\vec{a}| = \frac{|\vec{v}|^2}{|\vec{r}|}$$
$$\Leftrightarrow a = \frac{v^2}{r}$$
which is precisely the magnitude equation you gave. However, the direction of $\vec{a}$ must be towards the center, i.e. in the opposing direction of $\vec{r}$ which is produced by the minus sign.
