Show uniqueness of solution to $\dot{x}(t) = \frac{x}{2+2t}$ In a homework problem, the diff EQ came up 
$$\dot{x}(t) = \frac{x(t)}{2+2t}$$
and $x(1) = 1$. I note that $x(t) = \frac1{\sqrt{2}}\sqrt{1+t}$ is a solution. How might I show uniqueness of the solution?
I would say what I've tried, but I am inexperienced at diff EQ, and especially for equations which are functions of $x$ itself.
 A: As this is a separable equation, you can explicitly solve it and step-by-step, show that there are no other options. First assume that at some point in time, that $x\neq0$. (If $x=0$ for all points in time, then just note that is one solution to the equation but not to your initial condition.) Then the following is valid for all such $x$:
$$\frac{\dot{x}}{x}=\frac{1}{2+2t}$$
The right side is continuous for $t$ in $(-\infty,-1)$ and again for $t$ in $(-1,\infty)$. The two sides must have equal antiderivatives with respect to $t$ up to a constant on these regions where the left side is continuous:
$$\ln|x|=\begin{cases}\frac{1}{2}\ln|2+2t|+C_1&t<-1\\\frac{1}{2}\ln|2+2t|+C_2&t>-1\end{cases}$$
Given your initial condition $x(1)=1$, and simplifying absolute value,
$$\ln|x|=\begin{cases}\frac{1}{2}\ln(-2-2t)+C_1&t<-1\\\frac{1}{2}\ln(2+2t)-\ln(2)&t>-1\end{cases}$$
So 
$$x=\begin{cases}k\sqrt{-2-2t}&t<-1\\\frac12\sqrt{2+2t}&t>-1\end{cases}$$
And that is a definitive conclusion about $x$ in the neighborhood of $t=0$. Note the solution is not unique if you want to take it continuously into $(-\infty,-1]$.
A: If $x(t)$ and $y(t)$ are each solutions, then $x'(t)/x(t)=y'(t)/y(t)$ so that $yx'-xy'=0$. Note that if we define $z=x/y$ then the numerator of $z'$ is this term $yx'-xy'=0$ (denominator being $y^2$).
So we get that $z$ is constant, i.e. there is $c$ with $x(t)=cy(t).$ Now since $x(1)=y(1)=1$ we get $c=1$ so that $x(t)=y(t)$ for uniqueness.
Some care must be taken at steps where maybe a zero-divide. In particular we must restrict to subintervals where $x/y$ is defined and differentiable, in order to use "zero derivative implies constant".
A: For all $t_0>-1$, take $x_0 = \frac{1}{2}\sqrt{2+2t_0}$ as the initial condition. 
The Picard–Lindelöf theorem then guarantees uniqueness of the solution around some region $[t_0 - \epsilon, t_0 + \epsilon]$. 
