Suppose that $v$ is a distribution on the real line. Then under what conditions can I solve the differential equation

$$ \dot{x}(t) = v(x(t)) $$

which I might interpret as an integral equation

$$ -\int_{-\infty}^\infty x(t)\; \omega'(t)\; \mathrm{d}t = \int_{-\infty}^\infty v(x(t)) \; \omega(t)\;\mathrm{d}t $$

where $\omega$ ranges over compactly-supported test functions.

For instance, if $v(x) = \delta_1(x) + x$, then I would expect the following to be a solution

$$ x(t) = \begin{cases} e^t & \text{if $t < 1$} \\ (1 + e)e^{t-1} & \text{if $t > 1$} \end{cases} $$

Suppose that $v$ is a distribution on the real line. Then under what conditions can I solve the differential equation

$$ \dot{x}(t) = v(x(t)) $$

which I might interpret as an integral equation

$$ -\int_{-\infty}^\infty x(t)\; \omega'(t)\; \mathrm{d}t = \int_{-\infty}^\infty v(x(t)) \; \omega(t)\;\mathrm{d}t $$

where $\omega$ ranges over compactly-supported test functions.

For instance, if $v(x) = \delta_1(x) + x$, then I would expect the following to be a solution

$$ x(t) = \begin{cases} e^t & \text{if $t < 1$} \\ (1 + e)e^{t-1} & \text{if $t > 1$} \end{cases} $$

Edit:

Sorry, I've been confusing my $x$s and $t$s. It should be

$$ x(t) = \begin{cases} e^t & \text{if $t < 0$} \\ 2e^t & \text{if $t > 0$} \end{cases} $$

Suppose that $v$ is a distribution on the real line. Then under what conditions can I solve the differential equation

$$ \dot{x}(t) = v(x(t)) $$

which I might interpret as an integral equation

$$ -\int_{-\infty}^\infty x(t)\; \omega'(t)\; \mathrm{d}t = \int_{-\infty}^\infty v(x(t)) \; \omega(t)\;\mathrm{d}t $$

where $\omega$ ranges over compactly-supported test functions.

For instance, if $v(x) = \delta_1(x) + x$, then I would expect the following to be a solution

$$ x(t) = \begin{cases} e^t & \text{if $t < 1$} \\ (1 + e)e^t & \text{if $t > 1$} \end{cases} $$

Suppose that $v$ is a distribution on the real line. Then under what conditions can I solve the differential equation

$$ \dot{x}(t) = v(x(t)) $$

which I might interpret as an integral equation

$$ -\int_{-\infty}^\infty x(t)\; \omega'(t)\; \mathrm{d}t = \int_{-\infty}^\infty v(x(t)) \; \omega(t)\;\mathrm{d}t $$

where $\omega$ ranges over compactly-supported test functions.

For instance, if $v(x) = \delta_1(x) + x$, then I would expect the following to be a solution

$$ x(t) = \begin{cases} e^t & \text{if $t < 1$} \\ (1 + e)e^{t-1} & \text{if $t > 1$} \end{cases} $$