Trying to solve: $y'' + 2y' = \frac{(y')^2}{y + 1} + \frac{y'}{x} \ln(\frac{y+1}{y'})$, $y(1) = 1$, $y'(1) = 2/e$. The problem is the following:  Solve
$$y'' + 2y' = \frac{(y')^2}{(y + 1)} + \frac{y'}{x} \ln\left(\frac{y+1}{y'}\right),$$
given that $y(1) = 1$, $y'(1) = \frac{2}{e}$.
My solution:
After noting that
$$\left(\ln\left(\frac{y+1}{y'}\right)\right)' = \frac{y'}{y + 1} - \frac{y''}{y'}$$
we divide the original equation by $y'$, rearrange and get
$$\frac{y'}{y + 1} - \frac{y''}{y'} + \frac{1}{x}\ln\left(\frac{y + 1}{y'}\right) = 2$$
Then we substitute $t(x) = \ln\left(\frac{y + 1}{y'}\right)$. Now we have
$$t' + \frac{t}{x} = 2$$
Particular solution to this equation is $t = x$ and for $t' + \frac{t}{x} = 0$ we have $t = \frac{C}{x}$. So finally we have $t = x + \frac{C}{x}$.
Now we note that $t(1) = \ln\left(\frac{e(1 + 1)}{2}\right) = 1$ and conclude that $t = x$ and then solve for $y$.
For $y$ we have
$$\frac{y + 1}{y'} = e^x$$
Solution for this equation is $\ln|y + 1| = -e^{-x} + C$. After some manipulations we get
$$y = \frac{C}{e^{e^{-x}}} - 1$$
and following the initial condition $y(1) = 1$ we get $C = 2e^{e^{-1}}$. So the answer will be
$$y = 2e^{e^{-1} - e^{-x}}$$
I have 2 questions:
$\qquad 1)$ Is my solution correct? The answer looks terrible, so maybe I had an error somewhere.
$\qquad 2)$ Is there any easier solution? I thought about finding $y''(1)$, multiplying by $x/y'$, taking derivative to remove $\ln$-function and solve the resulting equation, but it seemed to be much harder than what I did in the solution above.

$$y'' + 2y' = \frac{(y')^2}{(y + 1)} + \frac{y'}{x} \ln\left(\frac{y+1}{y'}\right),$$
$$(\ln y')'+2=(\ln (y+1))'+\dfrac 1x (\ln( y+1))-\dfrac 1x \ln y'$$
$$x(\ln y')'+2x=x(\ln (y+1))'+ (\ln (y+1))- \ln y'$$
$$x(\ln y')'+\ln y'+2x=x(\ln (y+1))'+ \ln (y+1)$$
The integration is obvious.
$$x\ln y'+x^2=C+x \ln (y+1)$$
$$\ln (\dfrac {y'}{y+1})=\dfrac Cx-x$$
$$\ln ((\ln (y+1))')=\dfrac Cx-x$$
Your solution looks good. I see no mistakes.
