Does $x(t) = \exp\left(\frac{t}{t-1}\right)\cdot\theta(1-t)$ solve $\dot{x}\cdot(1-t)^2+x=0,\,\,x(0)=1$

Question

Does $x(t) = \exp\left(\frac{t}{t-1}\right)\cdot\theta(1-t)$ solve $\dot{x}\cdot(1-t)^2+x=0,\,\,x(0)=1$ with $\theta(t)$ the standard unitary step/Heaviside function

$$\theta(t) := \begin{cases} 0 & x\leq 0 \\ 1 & x > 0. \end{cases}$$

I know beforehand that $y(t) =\exp\left(\frac{t}{t-1}\right)$ solves $\dot{y}\cdot(1-t)^2+y=0,\,\,y(0)=1$, and I want to know if the same differential equation sustained as a finite-duration solution the cropped version $x(t)$.

If not, please explain which differential equation will be solved by $x(t)$ on the whole real line.

My attempts

Through Wolfram-Alpha I take the derivative of $x(t)$ and it looks it match the differential equation, except at one zero-measure point where a Dirac's Delta function rises, but because this term has the form $f(1-t)\delta(1-t)$ when multiplied by $(1-t)^2$ on the diff. eq., I believe lies under the property $x\cdot\delta(x) = 0$ so indeed $x(t)$ is solving the differential equation.

Since this system is not an autonoums ODE, it is different from the problems analyzed by V. T. Haimo: Finite Time Differential Equations and Finite Time Controllers, so I don't know how to apply here was is shown on the papers. Maybe someone with a better background on differential equations could explain how to figure out if indeed the differential equation stand a finite duration solution (these are the only related papers I have found so far).

But given I already know that $y(t)$ solves the irrestricted diff. eq., which also has the trivial solution $y(t)=0$, I believe that since $x(t)$ achieves zero "smoothly" there are no reasons to discard it as a solution… which is quite interestimg thinking about Uniqueness of solutions which should stand every Lipschitz ODE, so I guess the differential equation is not a Lipschitz ODE please confirm this explaining why.

Answer 1

Your equation is not an ODE at $t=1$, so as ODE it only has solutions for $t<1$ and for $t>1$.

However, as all these solutions have limit $0$ at $t=1$, you can combine any solution left of $t=1$ with any solution right of $t=1$ to get a continuous function that solves the given equation.

But no such combination is "The" solution of this equation. Your case is also such a combination, with the zero solution for $t>1$.