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?

• Here is a rigourous way to do the same thing \begin{align} y' &= xy \\ \implies y'/y &= x \\ \implies \int y'/y dx &= \int x dx \\ \implies \ln y &= x^{2}/2 + C \end{align} with the assumption $y \ne 0$. Else if $y = 0$ the DE is trivially satisfied. Commented Dec 11, 2019 at 16:03
• @mattos what about division by $y$? Commented Dec 11, 2019 at 16:04
• You can divide by y, as long as it is non zero.
– Gabe
Commented Dec 11, 2019 at 16:04
• This can be interesting for you: {math.stackexchange.com/q/21199.
– user
Commented Dec 11, 2019 at 16:07
• If one knows about differential forms, one is looking for integral curves of $\omega=0$ on $\Bbb R^2$, where $\omega = x\,dx - \frac 1y\,dy$. (This means we want curves whose tangent vector $v$ at the point $(x,y)$ satisfies $(x\,dx - \frac1y\,dy)(v) = 0$.) Commented Dec 11, 2019 at 18:45

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.