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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?

Michael Hardy
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Kishalay Sarkar
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    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. – Matthew Cassell Dec 11 '19 at 16:03
  • @mattos what about division by $y$? – Kishalay Sarkar Dec 11 '19 at 16:04
  • You can divide by y, as long as it is non zero. – Gabe Dec 11 '19 at 16:04
  • This can be interesting for you: {https://math.stackexchange.com/q/21199. – user Dec 11 '19 at 16:07
  • The geometrical significance of the ODE is that it gives a family of (kind of parallel curves called equi or iso, here it is $\ln y-x^2/2=C$, wehere $C$ takes real values. One may call $y$ as pressure $P$, $x$ as $V$ and $C$ as $T$. So these curves isothermals at $T_1,T_2,T_3,...$ – Z Ahmed Dec 11 '19 at 16:09
  • @Gabe what do you mean by $y$ non-zero,does it mean non-zero for some value or non-zero for all values. – Kishalay Sarkar Dec 11 '19 at 16:22
  • @user thanks for suggesting but my question is a bit different. – Kishalay Sarkar Dec 11 '19 at 16:23
  • @DrZafarAhmedDSc I am not looking for any physical meaning or analogy. – Kishalay Sarkar Dec 11 '19 at 16:24
  • Treat $dy/y = xdx$ for what it really is: just a simple rule for you to remember how to solve these kinds of equations. You don't need this method to be rigorous - you can easily differentiate and check the answer. As for dividing by $y$: this is perfectly fine as long as the initial value is not zero. If it is zero and the ODE does not have the uniqueness property then you have to be careful. One example of this is $y' = y^n$ with $y(0) = 0$ and $n<1$. Here there are infinitely many solutions other than $y=0$, see e.g. https://math.stackexchange.com/questions/519495/non-uniqueness-ode – Winther Dec 11 '19 at 16:25
  • When you use an interating factor you dont divide by y $\mu (x)=e^{-x^2/2}$ – user577215664 Dec 11 '19 at 16:26
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    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 $(x,dx - \frac1y,dy)(v) = 0$.) – Ted Shifrin Dec 11 '19 at 18:45
  • Sorry that you have to scavenge. –  Dec 11 '19 at 19:27

1 Answers1

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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.

Jam
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