Meaning of $\frac{1}{y}dy=x\,dx$. Consider the differential equation $\frac{dy}{dx}=xy$, If we go on mechanically we will follow the steps below:
$\frac{dy}{y}=x\,dx$,then integrate it to get $\ln(y)=x^2/2+C$. Fine, but what does $\frac{dy}{y}=x\,dx$ mean. Does it make any sense. Loosely speaking one can say $dy$ and $dx$ are infinitesimal change and all that rubbish things. But I want to understand is the any meaning of writing this or it is just a notation?Also there is another doubt, how can we divide both sides by $y$. I am looking for a rigorous understanding of these things. Can someone guide me?
 A: From an intuitive perspective, you can consider $\mathrm{d}x$ and $\mathrm{d}y$ as the small distances between two points, respectively $x_1$ and $x_2$ or $y_1$ and $y_2$. Then, we can algebraically express this as $\mathrm{d}x=x_2-x_1$ and, defining $y=f(x)$ as a function of $x$, we have $\mathrm{d}y=f(x_2)-f(x_1)$. So your prior expressions, treating $\mathrm{d}x,\mathrm{d}y$ as algebraic variables becomes $x\mathrm{d}x=x_2(x_2-x_1)$ or $x_1(x_2-x_1)$ and $\frac{\mathrm{d}y}{y}=\frac{f(x_2)-f(x_1)}{f(x_2)}$ or $\frac{f(x_2)-f(x_1)}{f(x_1)}$. When $x_2\approx x_1$ and $f(x_2)\approx f(x_1)$ the two expressions $x\mathrm{d}x$ effectively the same so it doesn't matter which one we pick and vice versa for $\frac{\mathrm{d}y}{y}$. Therefore the solution of the differential equation is the function that satisfies $\displaystyle x_1(x_2-x_1)\approx \frac{f(x_2)-f(x_1)}{f(x_1)}$ for any close $x_1, x_2$ that you pick. In other words, $\displaystyle {f(x+h)}\approx (1-hx)f(x)$, for small $h$. 
From a more rigorous perspective, we can argue that $\mathrm{d}x$ and $\mathrm{d}y$ are undefined terms, when given out of context. Therefore, the expression really states that $\frac{\mathrm{d}x}{\mathrm{d}y}=xy$. i.e. if $y=f(x)$ then, given any $x$, we have $f'(x)=xf(x)$. As Mattos stated in the comments, this can be solved with integration by substitution without ever having to separate the $\mathrm{d}x$ and $\mathrm{d}y$ into separate terms.
$$\begin{align}f'(x)&=xf(x)
\Rightarrow
\frac{f'(x)}{f(x)}=x
\\
\int\limits_{x_0}^{x} \frac{f'(t)}{f(t)}\,\mathrm{d}t&=\int\limits_{x_0}^{x} t\,\mathrm{d}t
\\
\text{LHS}&=\int\limits_{f(x_0)}^{f(x)} \frac{f' \circ f^{-1}(t)}{f\circ f^{-1}(t)}\cdot (f^{-1})'(t)\,\mathrm{d}t\tag{1}
\\
&=\int\limits_{f(x_0)}^{f(x)}\frac{\left[f' \circ f^{-1}(t)\right]\cdot \left[(f^{-1})'(t)\right]}{t}\,\mathrm{d}t
\\
\ln f(x) - \ln f(x_0)&=\frac{x^2}2-\frac{x_0^2}2\tag{2}
\\
f(x) &=\frac{f_0}{\exp\left(\frac{x_0^2}2\right)}\cdot \exp\left(\frac{x^2}2\right)
\end{align}$$
Equation $(1)$ uses integration by substitution, with $\varphi(x) = f^{-1}(x)$. Equation $(2)$ uses $\left[f' \circ f^{-1}(t)\right]\cdot \left[(f^{-1})'(t)\right]=1$, which can be proven with limits Question 315835.
