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$$\frac{1}{y} \frac{\text{d}y}{\text{d}x} = \frac{1}{x}$$

The way I solve this:

for all $x$ in $(-\infty,0)$

$\ln |y| = \ln |x| + A$ where $A$ is a real constant (since we know antiderivatives are separated by a constant)

$|y| = B|x|$ where $B$ is a positive constant equals $\exp(A)$

$y= Bx$ or $-Bx$

It seems to me $y =Cx$ where $C$ is a real non-zero constant that equals either $B$ or $-B$.

Since it seems like if $y$ is equal to $+$ or $-Bx$ for the entire interval $(-\infty,0)$, then it must be only $+Bx$ or $-Bx$ i.e. $Cx$ for some subinterval. If it is $+Bx$, then for values outside and near this interval, they must also be $+Bx$ since we know $y$ is continuous and $x$ is not zero. Vice versa.

But I don't know how to prove this. How to prove $y = Cx$ for $(-\infty, 0)$?

If that is the case, then I can write the general form of the differential equation picewisely as $y=C_1x$ for $(-\infty,0)$, $C_2x$ for $(0,\infty)$, where $C_1$ and $C_2$ are non-zero reals.

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  • $\begingroup$ If y=1 when x=1, then we know B = 1 and y = + or - x for x in (0,infinity). We also know y=x at x=1. By intuition, it then seems obvious that y=x for (0,inifinity), but how to prove this? $\endgroup$
    – TFR
    Dec 29, 2021 at 0:28
  • $\begingroup$ consider Intermediate Value Theorem $\endgroup$
    – TFR
    Dec 30, 2021 at 12:35

2 Answers 2

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The first point of confusion here is that you antidifferentiatied $$\frac{y'}{y}=\frac1{x}$$ to $$\ln(|y|)=\ln(|x|)+A,$$ but this is not entirely correct. The antiderivatives of $\frac1{t}$ are given by the piecewise, $$\ln(-t)+A;\,\forall{t\lt0}$$ $$\ln(t)+A;\,\forall{t\gt0}.$$ Taking this into account, you should have $$\ln(-y)=\ln(-x)+A$$ or $$\ln(y)=\ln(-x)+B.$$ This translates to $$y(x)=\exp(A)x$$ or $$y(x)=-\exp(B)x.$$

The second point of is in interpreting the "or." The two solutions we have here are that $y(x)=c_0x$ with $c_0\gt0,$ or that $y(x)=c_1x$ with $c_1\lt0.$ You could not have both happening simultaneously in all of $(-\infty,0).$ Your question is, how do you prove it? Well, notice that the equation must be satisfied everywhere. But it cannot be satisfied if there are jump discontinities, which occur if you have a constant of one kind in one subinterval, and a constant of some other kind everywhere else. You can show that if $$f(x)=\begin{cases}Ax&x\leq{C}\\Bx&x\gt{C}\end{cases},$$ then $f$ is not even differentiable at $C$ unless $A=B.$ You can prove it just using the definition of the derivative. The method for solving the equation, when done properly, already implies this, though, so in practice, you would not need to prove it.

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  • $\begingroup$ It’s good that you pointed out my first confusion about 1/x. For the proof of f(x) given, you just proved it cannot changes constants for two INTERVALS which seem to have been mentioned by me in the post. You seem to be assuming y have two coefficients for all negative x if and only if it does so for two sub-intervals. I don’t see how is that true. Could you expand on that? $\endgroup$
    – TFR
    Jan 8, 2022 at 13:26
  • $\begingroup$ I think a simpler and better way is to make use of the Intermediate Value Theorem from the start. Suppose y is positive for one negative x and sometimes negative for another negative x, then y is zero for some x between which is a contradiction. Therefore y is either positive for all negative x or negative for all negative x. And then an anti-derivative of LHS for all negative x would be ln(y), or ln(-y). $\endgroup$
    – TFR
    Jan 8, 2022 at 13:29
  • $\begingroup$ ln(−𝑦)=ln(−𝑥)+𝐴 ln ⁡ ( − y ) = ln ⁡ ( − x ) + A or ln(𝑦)=ln(−𝑥)+𝐵. $\endgroup$
    – TFR
    Jan 8, 2022 at 13:31
  • $\begingroup$ I don’t see how the above statement is arrived at. To find an anti-derivative of LHS for negative x, one requires substitution rule. Since y is sometimes positive and sometimes negative and never zero, then y is not continuous and substitution cannot be used. I don’t know how you did that. But I think it is unnecessary here as I already figured out a simple way as commented above. $\endgroup$
    – TFR
    Jan 8, 2022 at 15:28
  • $\begingroup$ I will green tick if you could clarify my two doubts above, as, again, the first point is very useful. $\endgroup$
    – TFR
    Jan 8, 2022 at 15:32
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Consider the DE in its explicit normal form $\frac{dy}{dx}=\frac{y}{x}$. The difference is that the equation is now defined for $y=0$, inside the quadrants the solutions remain the same.

As an ODE, this equation is singular or not defined on the line $x=0$. Thus the solutions on $(-\infty,0)$ and $(0,+\infty)$ are separate, with separate integration constants. For the solution as ODE one need not care much of how they connect at $x=0$.

$y=0$ is a solution of the explicit ODE, or outside the domain for the original equation. Thus no other solution can change its sign, trivially by not being able to leave its quadrant, or by the uniqueness property of the Picard-Lindelöf (-Cauchy-Lipschitz) theorem on existence and uniqueness. As the sign is fixed, it is determined by the initial condition, one could write $$ \frac{x}{x_0}=\frac{y}{y_0}. $$

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  • $\begingroup$ If you put the DE in normal form $y'=y/x$, as one usually does for solving DE. $\endgroup$ Dec 30, 2021 at 9:04
  • $\begingroup$ This does not properly address the question. $\endgroup$
    – Angel
    Jan 3, 2022 at 14:10
  • $\begingroup$ @Angel: In what way? The question was asking why the constant factor in the solution has a constant sign, the answer is that no sign change takes place in any component, thus the signed constant is completely determined by the initial condition. $\endgroup$ Jan 3, 2022 at 14:16
  • $\begingroup$ I assume you use “change of sign” to mean the coefficient of y is constant for all negative x (another constant for all positive x) (what I was originally asking for a proof about). I think you can just see no solution y can “changes its sign” by simply noting 1/y and y is continuous. There is no need for all of that, right? Sorry if I interpret it wrong, what you did seem to only prove y is never zero using the uniqueness property, rather than by simply noting 1/y. Also you did not explain explicitly how y not equal to zero means it cannot “change sign”. $\endgroup$
    – TFR
    Jan 8, 2022 at 16:04
  • $\begingroup$ I green ticked this question because when I read this, I considered y, for negative x, changes its sign only if it is zero for some x, leading to a contradiction, and therefore the coefficient of y is constant for all negative x. $\endgroup$
    – TFR
    Jan 8, 2022 at 16:05

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