How can I solve this non-linear first order differential equation?

[DryRun]

Homework Statement
http://s1.ipicture.ru/uploads/20120128/U1SBMa6O.jpg

The attempt at a solution
I expanded the equation and got:
$$\frac{dy}{dx}=\frac{ylny-ylnx+y}{x}$$
I can't use the method of separation of variables, nor is this a homogeneous equation. It's not a 1st order linear ODE nor in the Bernoulli format. I'm stuck.

 

[ehild]

Hi sharks,
The right side of the equation is function of y/x. Try new variable y/x=u.

ehild

 

[DryRun]

Hi ehild!

So, i guess this is a homogeneous equation.
$$\frac{dy}{dx}=u(lny-lnx+1)$$
I'm supposed to eliminate all y and x on the R.H.S. and replace by u only.
$$\frac{dy}{dx}=u(lny-lnx+1)=u(lnu+1)=ulnu+u$$
From $y=ux$, $\frac{dy}{dx}=u+x\frac{du}{dx}$. Then, $u+x\frac{du}{dx}=ulnu+u$.

So, $x\frac{du}{dx}=ulnu$
$$\frac{dx}{x}=\frac{1}{ulnu}\,.du$$

 

[ehild]

And what is dy/dx in terms of u and x?

ehild

 

[DryRun]

I have edited my previous post but then I'm stuck at integrating the R.H.S. of: $$\int \frac{dx}{x}=\int \frac{1}{ulnu}\,.du$$
$$lnx=\int \frac{1}{ulnu}\,.du$$
If i consider integrating $1/u$ then i get $lnu$.

 

[ehild]

1/u is the differential of ln(u). So the integrand is of form 1/ln(u) (lnu)'
I just can not understand what you did.

ehild

 

[DryRun]

The integration result would then be $ln(lnu)+lnA$, where lnA is the arbitrary real constant of integration.
Therefore, $lnx=ln(lnu)+lnA$ or $x=Alnu$ or $\frac{x}{A}=lnu$
$$\frac{x}{A}=ln\frac{y}{x}$$
$$\huge e^\frac{x}{A}=\frac{y}{x}$$
So, the final answer is (i'm using big font to make the power fraction more visible):
$$\huge y=xe^\frac{x}{A}$$
Is that correct?

 

[ehild]

Try to substitute it back into the original equation.

ehild

 

[DryRun]

Yes, i get the same L.H.S and R.H.S. that is,
$$\huge \frac{xe^\frac{x}{A}}{A}+e^\frac{x}{A}$$
So, i assume this is the proof. Thanks, ehild.

 

[ehild]

You are welcome Remember that y/x substitution.

ehild