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I have to solve $\dfrac{dy}{dx} = \dfrac{y}{x}$. So I set $v = \dfrac{y}{x}$ and so $$ \dfrac{dy}{dx} = v $$ Then by product rule $x\dfrac{dv}{dx} + v = v$ and so $x\dfrac{dv}{dx} = 0$. But then that means there is no unique solution to the differential equation; am I wrong in my reasoning?

Wolframalpha said the solution was $y(x) = cx$.

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  • $\begingroup$ Since $\frac{dv}{dx}=0$, you know that $v(x)=c$ for all x, for some constant c. Now you can substitute back for v. $\endgroup$ – user84413 Sep 4 '13 at 0:46
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Approach 1

$$\dfrac{dy}{y} = \dfrac{dx}{x}$$

Integrate each side and simplify, you get:

$$y = c x$$

Approach 2

This is your approach, we get:

$$v' x = 0 \rightarrow v = c$$

From the initial substitution, you have:

$$v = \dfrac{y}{x} \rightarrow y = v x = cx$$

In other words, both methods yield the same family of curves and your reasoning is correct.

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  • $\begingroup$ I'm not trying to solve it with this method. Why does the other method fail? $\endgroup$ – Don Larynx Sep 4 '13 at 0:36
  • $\begingroup$ See the update. $\endgroup$ – Amzoti Sep 4 '13 at 0:47
  • $\begingroup$ Nice work, Amzoti! $\endgroup$ – Namaste Sep 4 '13 at 0:48
  • $\begingroup$ Neither method fails. And they both give the same family of answers. $\endgroup$ – André Nicolas Sep 4 '13 at 0:50
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    $\begingroup$ @Ovi: you are very welcome! Regards $\endgroup$ – Amzoti Sep 4 '13 at 22:32
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${\frac{dy}{dx}}={\frac{y}{x}}$

${\frac{dy}{y}}={\frac{dx}{x}}$

Integrating

${\ln}y={\ln}x+C$,

where $C$ is an integration constant.

${\ln}y={\ln}x+{\ln}a$

${\ln}y={\ln}(ax)$

$y=ax$

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  • $\begingroup$ This doesn't actually answer the question, which was "am I wrong in my reasoning?" $\endgroup$ – 6005 Sep 4 '13 at 0:53

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