Is $\frac{1}{x} x$ defined when $x = 0$? My professor was teaching us differential equations where he tried to solve the equation 
$$y' +\frac{y}{x} =0$$
After obtaining a general solution he asked whether the solution exists for x=0? The natural response was no. 
But then he transformed the equation into 
$$xy'+y=0$$ saying that it should exist. 
My question is "Is such a manipulation allowed ?" Considering the new transformation doesn't allow the integrating factor method, moreover the transformation was done assuming $\frac{1}{x} x$ is defined when $x = 0.$ 
 A: You're approaching the problem from the wrong side, I believe. Anyway, your objection is valid, although for a slightly different reason.
Suppose you have a solution $f(x)$ of $y'+y/x=0$; a natural question to ask is whether such a function can be extended to a continuous function at $0$. Asking whether such extension is a solution of the differential equation at $x=0$ does not apply (differential equations are solved over intervals, not at points).
Now, since for $x\ne0$, you have $xf'(x)+f(x)=0$, you can draw some conclusions, at least in some cases. The general solution of the equation is
$$
f(x)=\begin{cases}
c_+/x & x>0 \\[4px]
c_-/x & x<0
\end{cases}
$$
where $c_+$ and $c_-$ are arbitrary constants. There's no way to extend such a solution to $0$ so that the resulting function is continuous at $0$, except for the trivial one when $c_+=c_-=0$.
If the equation is instead $y'-y/x=0$, then the situation is quite different:  all solutions of this equation can be extended to $0$. Indeed, the general solution has the form
$$
f(x)=\begin{cases}
c_+x & x>0 \\[4px]
c_-x & x<0
\end{cases}
$$
where $c_+$ and $c_-$ are arbitrary constants.
Some of the solutions are even differentiable at $0$ (precisely, those for which $c_+=c_-$); others aren't.
For this equation, $x=0$ is a singular point; solutions need not be extendable at the singular points even if algebraic manipulations remove the denominator. Think to $y'+y/x^2=0$, which can be rewritten as $x^2y'+y=0$, but whose solutions (except the constant one) cannot be extended by continuity at $0$.
Thus it's not sufficient to rewrite the equation without the fraction to guarantee that the solutions can be extended by continuity at $0$. More generally, if you have an equation such as $A(x)y'+B(x)y=0$, where $A(x)$ and $B(x)$ are defined and continuous on some open set, the zeros of $A(x)$ in this domain will be singular points and there's no general theorem that guarantees the solutions can be extended by continuity at these singular points.
A: Three original equation isn't defined for $x=0$, you're right. However, you can't (at least on general principles) rule out the possibility of differentiable functions $f:\Bbb R\to\Bbb R$ that fulfill the original equation, and are defined at $x=0$. Such a function will also fulfill the other equation. 
In other words, he transformed the differential equation into another that isn't entirely equivalent, but is satisfied by the same functions, so for all intents and purposes they are interchangeable.
A: No,$\dfrac{1}{x}\times x$ is not defined for $x=0$ as $\dfrac{1}{0}$ is not defined. 
Hope this helps
