I want to study this differential equation, that is to justify the unique maximal solution and determine the interval where the solution holds (if it exists).
We have:
$$\left\{\begin{matrix} y' =\frac{xy}{(x-1)^2} \\ y(2)= 1 \end{matrix}\right.$$
My first attempt is to try to figure out where could the solutions work. I want to use the Picard–Lindelöf theorem.
Let $f(t,y) = \frac{ty}{(t-1)^2}$. We need to verify that this function is continuous on the interval of definition and that it is k-Lipschitz. $f$ is continuous only on two intervals: $I_1 = (-\infty, 1)$ and $(1, +\infty)$. I can't prove that it is k-Lipschitz on the intervals. So let's take $[a,b] \subset I_2$ such that $2 \in [a,b]$. We have:
$$|f(t,y) - f(t,y)| = | {\frac{t(y-g)}{(t-1)^2}}| \leq |\frac{b(y-g)}{(t-1)^2}|$$ Let $m = \min \{ (x-1)^2 | x\in [a,b] \}$, we then have:
$$|\frac{b(y-g)}{(t-1)^2}| \leq |\frac{b(y-g)}{m}| = \frac{b}{m}|y-g|$$ Thus it is k-Lipshitz.
So it satisfies the hypotheses of Picard–Lindelöf theorem.
Now, to solve it, I just put $f(t,y) = h(y)g(x)$ with $h(y) = y$ and $g(x) \frac{x}{(x-1)^2}$.
And the result is given by: $$ \int _{y(t)}^{y(t_0) = 1} \frac{dz}{h(z)} = \int _{t}^{t_0=2}g(x)dx$$
Is my reasoning sufficient? Do I need to do something more? Any hints, remarks, criticism are greatly appreciated!
Note that $$ \int \frac {x}{(x-1)^2}dx = \int \frac{u+1}{u^2} du = \ln |x-1| - \frac {1}{x-1} +C$$
Solving the separable equation and considering the initial condition results in $$y=(x-1)e^\frac{x-2}{x-1}$$