this was a problem from a textbook:

If $x>0$, $y>0$, find the general solution to the differential equation, $$ x \frac{dy}{dx} = y + \frac{x}{\ln y - \ln x }$$ giving your answer in the form $ye^{y/x}=f(x)$

I approached with the substitution $y=vx$ since this is a homogenous equation. Giving,

$$ x\frac{dv}{dx}+v = v+ \frac{1}{\ln v}$$

By separating variables yields,

$$ v \ln v -v = \ln x + C $$

The by taking the power of $e$ $$e^{v \ln v - v } = xA$$

where $e^C= A$. This is equivalent to

$$e^{-v}v^{v} = Ax$$

Now when I substitute the value of $v= \frac{y}{x}$ back, I cannot obtain the equation of the form $ye^{y/x} = f(x)$.

Maybe I have done a mistake in my calculations. May someone explain where? Or how to solve the problem? Thank you so much!!

  • 1
    $\begingroup$ Everything looks fine to me in your solution. $\endgroup$ – Start wearing purple Apr 23 '15 at 14:23
  • $\begingroup$ The problem is I cannot manipulate the equation back to the desired form of the problem: $ye^{y/x} = f(x)$ $\endgroup$ – CL. Apr 23 '15 at 14:35
  • $\begingroup$ Maybe you can use the Lambert W function, but I don't know all its properties ..:) $\endgroup$ – Emilio Novati Apr 24 '15 at 16:19

Here my solution ( maybe someone can find something more simple).

Using the substitution $\log y-\log x=t \iff y=xe^t$ the differential equation become: $$ te^t\,dt=\dfrac{dx}{x} $$ with solution: $$ e^t(t-1)=\log x +c $$ divide by $e$: $$ (t-1)e^{t-1}=\dfrac{\log x+c}{e} $$ and using the Lambert W function we find: $$ t-1=W\left(\dfrac{\log x+c}{e}\right) $$ so we have (see here) $$ e^t=\dfrac{y}{x}=\dfrac{\log x +c}{W\left(\dfrac{\log x+c}{e}\right)}=e\dfrac{\dfrac{\log x +c}{e}}{W\left(\dfrac{\log x+c}{e}\right)}= $$ $$ =ee^{W\left(\dfrac{\log x+c}{e}\right)}=F(x) $$ and $y=xF(x)$ is the solution of the differential equation.

If we want the form requested in OP we can do: $$ ye^{\frac{y}{x}}=xF(x)e^{F(x)} $$


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