Non- linear ODE's with no independent variable I was going through some notes of a course on ODE's. I stumbled upon these equations:
$$y=y' \ln y'$$
$$y' =  e^{\frac {y'} y}$$
$$y=(y'-1)e^{y'}$$ 
I tried making $u=y'$ But then they wouldn't be differential equations, I also tried finding a parametric solution using the same substitution. Is there a method of solution I'm not aware of?
 A: The first one:
$$\begin{array}{rcl}
y &=& y' \ln y' \\
W(y) &=& \ln y' \\
y' &=& e^{W(y)} \\
\dfrac{\mathrm dy}{\mathrm dx} &=& e^{W(y)} \\
\dfrac{\mathrm dx}{\mathrm dy} &=& e^{-W(y)} \\
x &=& \displaystyle \int e^{-W(y)} \ \mathrm dy \\
&=& \displaystyle \int e^{-u} \ \mathrm d(ue^u) \\
&=& \displaystyle \int (1+u) \ \mathrm du \\
&=& \displaystyle u+\dfrac12u^2 + C \\
&=& \displaystyle W(y)+\dfrac12W(y)^2 + C \\
W(y)^2 + 2W(y) - (2x+C) &=& 0 \\
W(y) &=& \dfrac{-2\pm\sqrt{4+8x+4C}}{2} \\
W(y) &=& -1\pm\sqrt{1+2x+C} \\
W(y)e^{W(y)} &=& (-1\pm\sqrt{1+2x+C})e^{-1\pm\sqrt{1+2x+C}} \\
y &=& (-1\pm\sqrt{1+2x+C})e^{-1\pm\sqrt{1+2x+C}}
\end{array}$$

The second one:
$$\begin{array}{rcl}
y' &=&  e^{\frac {y'} y} \\
\ln y' &=& \dfrac {y'}y \\
y &=& \dfrac{y'}{\ln y'} \\
f(y) &=& \ln y' \\
y' &=& e^{f(y)} \\
\dfrac{\mathrm dy}{\mathrm dx} &=& e^{f(y)} \\
\dfrac{\mathrm dx}{\mathrm dy} &=& e^{-f(y)} \\
x &=& \displaystyle \int e^{-f(y)} \ \mathrm dy \\
&=& \displaystyle \int e^{-u} \ \mathrm d(e^u/u) \\
&=& \displaystyle \int \left(1-\dfrac1{u^2}\right) \ \mathrm du \\
&=& \displaystyle u + \dfrac1u + C \\
&=& \displaystyle f(y) + \dfrac1{f(y)} + C \\
[f(y)]^2 - (x+C)f(y) + 1 &=& 0 \\
f(y) &=& \dfrac{x+C\pm\sqrt{(x+C)^2-4}}2 \\
y &=& \dfrac{\exp \left(\dfrac{x+C\pm\sqrt{(x+C)^2-4}}2\right)}{\dfrac{x+C\pm\sqrt{(x+C)^2-4}}2}
\end{array}$$

The third one:
$$\begin{array}{rcl}
y &=& (y'-1)e^{y'} \\
f(y) &=& y' \\
\dfrac{\mathrm dy}{\mathrm dx} &=& f(y) \\
\dfrac{\mathrm dx}{\mathrm dy} &=& \dfrac1{f(y)} \\
x &=& \displaystyle \int \dfrac1{f(y)} \ \mathrm dy \\
&=& \displaystyle \int \dfrac1u \ \mathrm d\left((u-1)e^u\right) \\
&=& \displaystyle \int \dfrac{e^u+(u-1)e^u}u \ \mathrm du \\
&=& \displaystyle \int \dfrac{ue^u}u \ \mathrm du \\
&=& \displaystyle \int e^u \ \mathrm du \\
&=& e^u + C \\
&=& e^{f(y)} + C \\
f(y) &=& \ln(x-C) \\
(f(y)-1)e^{f(y)} &=& \left[\ln(x-C)-1\right]\left(x-C\right) \\
y &=& \left[\ln(x-C)-1\right]\left(x-C\right) \\
\end{array}$$
