# Why is it legitimate to solve the differential equation $\frac{dy}{dx}=\frac{y}{x}$ by taking $\int \frac{1}{y}\ dy=\int \frac{1}{x}\ dx$?

Answers to this question Homogeneous differential equation $\frac{dy}{dx} = \frac{y}{x}$ solution? assert that to find a solution to the differential equation $$\dfrac{dy}{dx} = \dfrac{y}{x}$$ we may rearrange and integrate $$\int \frac{1}{y}\ dy=\int \frac{1}{x}\ dx.$$ If we perform the integration we get $\log y=\log x+c$ or $$y=kx$$ for constants $c,k \in \mathbb{R}$. I've seen others use methods like this before too, but I'm unsure why it works.

Question: Why is it legitimate to solve the differential equation in this way?

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@Amzoti I don't think your link explains why the method is valid. In particular, I think the OP is asking what $dy/y = dx/x$ means. (And if it has no formal meaning, how can the argument be valid?) –  Trevor Wilson Sep 4 '13 at 1:15
Yes the ODE. In other words then, why is it legitimate to solve separable ODEs in this manner? Or perhaps, why does it work? We can manipulate the equation in all sorts of ways, most of which won't be useful; why do this? –  Rebecca J. Stones Sep 4 '13 at 1:15
@Rebecca In short, this calculational technique is the u-substitution theorem which more or less is the integration-version of the chain rule. Ultimately, the legitimacy is evidenced by the success of the method. –  James S. Cook Sep 4 '13 at 1:15
@TrevorWilson: Fair enough, she can look at math.bd.psu.edu/faculty/jprevite/251f11/250bookSec2.1.pdf for the actual proof of the method. –  Amzoti Sep 4 '13 at 1:17
It is justified by the chain rule –  oldrinb Sep 4 '13 at 12:43

You start with $$y'=\frac{y}{x}\implies \frac{y'}{y}=\frac{1}{x}\implies\int\frac{y'dx}{y}=\int \frac{dx}{x},$$ and you make the change of variables in the first integral, which results in what you've written $$\int\frac{dy}{y}=\int \frac{dx}{x}$$

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Thanks! I quite like this answer in particular, as it highlights the change of variables (which seems to be the crux of the technique). Thanks for the other answers too! –  Rebecca J. Stones Sep 4 '13 at 1:34

To be honest I think it's BS to teach separable variables like this without the Riemann-Stieljes integral.
The way I solve them is by doing what actually is done: integrate with respect to $x$ on both sides.

Remember that $y$ is a function (on the variable $x$). So your differential equation is, for all $x$ in a certain interval, $y'(x)=\dfrac{y(x)}{x}$ or equivalently $\dfrac{y'(x)}{y(x)}=\dfrac {1}{x}$and integrating with respect to $x$ you get the desired result.

In my opinion integrating with respect to $y$ is nothing more than a cheap trick, the same way $\dfrac{dy}{dx}=1\iff dy=dx$ is a cheap trick. It works only because of some higher math.

More generally, if you can rewrite your DE as $g(y(x))y'(x)=f(x)$ for some functions $f$ and $g$ that have antiderivatives, $F$ and $G$, in the given interval, then $g(y(x))y'(x)=f(x)\iff G(y(x))=F(x)+C$, for some $C\in \Bbb R$. (To establish $\Longleftarrow$ just differentiate). And if we're lucky enough for $G$ to be invertible, we get $y(x)=G^{-1}\left(F(x)+C\right)$. If $G$ isn't invertible, then hopefully the implicit function theorem will yield the solutions to the DE implicitly by the equation $G(y(x))=F(x)+C$.

In your example $g$ is the function $t\mapsto \dfrac{1}{t}$ which has $t\to \log (|t|)$ as an antiderivative. (Don't forget the absolute value).

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I agree, and this caused me a lot of confusion when I was first learning ODEs. –  littleO Sep 4 '13 at 1:20
+1 Nicely done, Git –  Babak S. Sep 4 '13 at 1:21
Why is measure theory needed here? Generally the problems encountered in a first course in ODEs have functions nice enough that the Riemann integral suffices to handle them. –  Potato Sep 4 '13 at 2:28
But in any case, as @GitGud showed in his answer, we can solve this ODE just fine without using Riemann-Stieltjes integration or anything fancy. (In fact, we only need to know that if $f' = g'$ on some open integral, then $f$ and $g$ differ by a constant on that interval.) –  littleO Sep 4 '13 at 20:41
No problem. Nice answer, by the way. –  Potato Sep 4 '13 at 21:48

I don't like the notation that's often used when solving ODEs. I'd prefer to write the solution like this:

\begin{align} & y'(x) = \frac{y(x)}{x} \quad \text{for all }x > 0 \\ \implies & \frac{y'(x)}{y(x)} = \frac{1}{x} \quad \text{for all }x > 0 \\ \implies & \log y(x) = \log x + C \quad \text{for all } x > 0 \,(\text{for some } C \in \mathbb R). \end{align}

(I'm assuming $y(x) > 0$ for all $x>0$.)

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