In a separable first-order ODE, we have the following form :
$y' = g(x)h(y)$
It follows that $h(y) = 0$ poses a a set of solutions, potentially, to the ODE.
Now, understanding that concept, I came across the following example where I have to find the general solution to the following ODE given that $ y(1) = 1 $
$ y' = \frac{(1+x)y}{x^2} $
Solving $ h(y) = 0$ gives the solution $y = 0$ however, this has been said to be 'incompatible' with $y(1) = 1$
Any further explanation that may help my understanding would be highly appreciated!
Don't get too lost in the wording. The explanation is
$ y(x) = 0 $ implies $y(1) = 0$. However, the initial condition given is $y(1)=1$.
Therefore it can't be a solution. Simple as that.