If $x>0$ and $y>0$ find the general solution to the differential equation giving your answer in the form $ye^{\frac{y}{x}}=f(x)$

Question

The Differential Equation is $xy'=y+\frac{x}{lny-lnx}$

This turns into a homogenous equation that takes the form $y'=\frac{y}{x}+\frac{1}{ln(\frac{y}{x})}$

Using the substitution $y=vx$ and $y'=xv'+v$

It turns into $xv'+v=v+\frac{1}{lnv}$

Which is now a separable differential equation which can be solved through

$\int lnv \, dv = \int \frac{1}{x} \, dx$

Which gives

$vlnv-v=lnx+c$

Since $y>0$ and $x>0$ absolute values are not needed

Raising both sides to the power of $e$ gives

$\frac{v^v}{e^v}=Ax$ Where $A=e^c$

Replacing $v$ with $\frac{y}{x}$

We get $\frac{\frac{y}{x}^\frac{y}{x}}{e^\frac{y}{x}}=Ax$

At this point I have no idea how to proceed because I don't see any way of getting it into the form of $ye^{\frac{y}{x}}=f(x)$. Either I have done something wrong prior to finding this or there is some method of doing it that I am not seeing.

For reference the answer is given as $ye^\frac{y}{x}=Ax^2$

Answer 1

The proposed solution would give you calculating backwards $$ v+\ln v=\ln A+\ln x\implies v'\left(1+\frac1v\right)=\frac1{x}\iff xy'=y+\frac{xy}{x+y} $$ which does not look close to your given ODE. I see no error in your solution, either the task is wrong or something is wrong between the assignment of tasks and reference solutions.