Solve $2x''\ln (x') =x' \,\, x(0)=1, x'(0)=e$.
My attempt
$p=x' \implies x''=pp'.$
$2pp'\ln p=p$.
$p(2p' \ln p-1)=0$.
We have that $p=0 \implies x(t)=c$ and also that $2p'\ln(p)=1 \implies \int \ln(p)dp=\frac 12 \int dx$.
$p\ln (p)-p=\frac 12x+c$.
In the picture you can See that I try to write in a different form
On
Note that we have $$2x''\ln (x') =x' \,\, x(0)=1, x'(0)=e$$ $$2\frac {x''}{x' }\ln (x')=1$$ Substitute $\ln x' =w$ $$\implies 2w'w=1 \implies (w^2)'=1$$ Can you take it from there ?
On
As I said, if $x'(t)=p(t)$, then $x''(t)=p'(t)$. So the equation can be written as $$2p'\ln(p)=p,$$ and you can separate variables and integrate as in $$2\int\frac{\ln(p)}p\,dp=\int dt.$$
Once you integrate use the fact that $x'(0)=p(0)=e$ and once you solve for $p(t)=x'(t)$ integrate once more to solve for $x(t)$ and use the condition $x(0)=1$.
Let $u=x'$. The equation can be written as $$ 2\frac{u'}{u} \ln u=1, $$ that is a separable equation, leading to $$ 2 \int \frac{\ln u}{u} du = t + C_1 $$ This integral can be evaluated using integration by parts identifying $U=\ln u$ and $dV = du/u$, leading to $$ (\ln u)^2= t + C_1 $$ $$ u = \exp \sqrt{t + C_1} $$ We have $x'(0)=u(0)=e$, then $C_1=1$. $x$ is given by $$ x = C_2+\int u dt = C_2 + \int \exp \sqrt{t+1 } dt, $$ This integration can be evaluated with the substitution $v=\sqrt{t+1}$, leading to $dt=2vdv$. The integral is now $2\int v \ \exp v \ dv$, which can be evaluated using integration by parts, leading to $$ x = C_2+ 2 \left(\sqrt{t+1}-1 \right) \exp \sqrt{t+1} $$ Using $x(0)=1$ we have $C_2=1$. Therefore, the solution is $$ x = 1 + 2 \left(\sqrt{t+1}-1 \right) \exp \sqrt{t+1} $$