# Homogeneous differential equation $\frac{dy}{dx} = \frac{y}{x}$ solution?

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 $y(x) = cx$.

• 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. – user84413 Sep 4 '13 at 0:46

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.

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

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

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

Integrating

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

where $C$ is an integration constant.

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

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

$y=ax$

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