Differential Equation - Where does the solution end? I was asked to solve the differential equation $y'+\frac{y}{x+1}=\frac{2y-1}{x}$, given the starting point y(0.5)=5/6. The equation meets the criteria for Existence and Uniqueness for every x>0 (as y' is continuous and so is dy'/dy). 
The solution to this problem is y=$\frac{2x^2+2x+1}{2x+2}$.
I need to answer where the solution exists and if it's continuously differentiable. Obviously the solution continues normally for every x>0, but what happens at x=0? Does the solution end there? Does it end in x=-1? Does it continue normally or is something else happening(like intersection of solutions outside of Uniqueness area)?
Is it continuously differentiable at x=0? and if so is there another problematic point?
Thanks in advance. 
 A: Your initial $x$ value is $0.5>0$ so the range of $x$ values where your solution is valid is $x>0$. If you are looking for a continuous solution, then you have to stop there since you can't jump past $0$.
The reason is that $x=-1,0$ are not allowed for the way your differential equation is written. Even though your solution doesn't have a problem at $x=0$, it is still not a solution for your differential equation at $x=0$ because your ODE can't handle $x=0$.
There are examples where the "bad value" of $x$ is not evident from the differential equation itself though. Usually these occur where the derivative goes vertical.
Note that $\frac{2y-1}{x}=\frac{4x+2}{2x+2}$ so we can write a new ODE:
$$y'+\frac{y}{x+1}=\frac{4x+2}{2x+2}$$
which has the exact same solution except that it is allowed to go past $x=0$, so the solution is valid here for $x>-1$.
A: The differential equation, as written, is not defined at $x=0$.  
The solution $y = (2x^2 + 2x + 1)/(2x + 2)$ exists and is continuously differentiable for $x > -1$, but it is not a solution of the differential equation at $x=0$ because that differential equation is not defined there.
