Log jammin' DE

Let's us differentiate both sides of our DE with respect to $x$ such that:

$\frac{d}{dx} [y' = ln(xy' - y)] \Rightarrow y'' = \frac{(y' + xy'') - y'}{xy' - y} = \frac{xy''}{xy' - y}.$

This in turn results in $\frac{x}{xy' - y} = 1 \Rightarrow y' - \frac{1}{x}y = 1$ , which after introducing the integrating factor $u(x) = \frac{1}{x}$ now produces:

$\frac{1}{x}y' - \frac{1}{x^{2}}y = \frac{1}{x} \Rightarrow (\frac{y}{x})' = \frac{1}{x} \Rightarrow \frac{y}{x} = ln(x) + C$

which after implementing the given boundary condition yields $\frac{-1}{1} = ln(1) + C \Rightarrow C = -1,$ or $y(x) = xln(x) - x.$

We wish to compute $y(e)$ , which equals $y(e) = e[ln(e) - 1] = e(1 - 1) = \boxed{0}.$

We want to get rid of the $\text{ln}(x)$ -function first, so we build two equations by applying the $\text{exp}(x)$ -function and by differentiation:

$\boxed{e^{y'}=xy'-y}$

$\boxed{y''=\frac{1}{xy'-y}}$

By putting those two equations together we get: $y''=\frac{1}{e^{y'}}=e^{-y'}$ .

Now substitute $y'=v$ and get: $v'=e^{-v}$ . This can easily be solved: $\frac{\text{d}v}{\text{d}x}=e^{-v} \Leftrightarrow \int e^v \text{d}v=\int \text{d}x \Rightarrow e^v=\boxed{e^{y'}=x}$

Substitute this result into our equation above: $y''=\frac{1}{e^{y'}}=\frac{1}{x}$

Now solve this by integrating twice:

$y''=\frac{1}{x} \Rightarrow y'=\text{ln}(x)+c_1 \Rightarrow \boxed{y=x(\text{ln}(x)-1)+c_2}$

Now apply the initial condition $y(1) = -1$ and get: $y(1)=-1=1(0-1)+c_2=-1+c_2 \Rightarrow \boxed{c_2=0}$

At last we can solve the problem: $\boxed{y(e)=e(1-1)=0}$