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$$2x''\ln (x') =x' \,\, x(0)=1, x'(0)=e$$ 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 $$2p'\ln(p)=1$$ $$\implies \int \ln(p)dp=\frac 12 \int dx$$ $$p\ln (p)-p=\frac 12x+c$$

![enter image description here](https://i.stack.imgur.com/0k6Z2.jpg]

In the picture you can See that I try to write in a different form

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  • $\begingroup$ I hope I didn't made mistakes .. $\endgroup$ – Isham Nov 11 '18 at 17:24
  • $\begingroup$ If $p(t)=x'(t)$, then $x''(t)=p'(t)$... Note that $x''(t)$ is the derivative of $x'(t)$, that is, the derivative of $p(t)$. $\endgroup$ – Alejandro Nasif Salum Nov 11 '18 at 17:30
  • $\begingroup$ @AlejandroNasifSalum p' is according to x not t $\endgroup$ – Isham Nov 11 '18 at 17:39
  • $\begingroup$ You mean $p'=\frac{dp}{dx}$? $\endgroup$ – Alejandro Nasif Salum Nov 11 '18 at 17:43
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    $\begingroup$ @AlejandroNasifSalum : Yes, the method is to assume that along a solution curve segment $x$ where $x$ is monotonous one re-parameterizes the curve $(t,x(t),x'(t))$ in phase space by $x$, so $t=s(x)$ and $x'=p(x)$. Then try to find a functional expression for $p$. $\endgroup$ – LutzL Nov 11 '18 at 18:29
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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 ?

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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 $x(0)=1$.

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How did you get $$p=x′⟹x′′=pp′$$

If you have $p=x'$ you will get $x''=p'$ why do you have an extra $p$ ?

Please fix that error and do your problem again.

The substitution is good just be careful.

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  • $\begingroup$ Thats correct Mohammaed the derivative p' is according to x not t so we have $p'p=x''$ $\endgroup$ – Isham Nov 11 '18 at 17:38
  • $\begingroup$ @Isham Are you sure? Think about it again. $\endgroup$ – Mohammad Riazi-Kermani Nov 11 '18 at 17:43
  • $\begingroup$ Mohammed try it yourself $$x''=\frac {dp}{dt}=\frac {dp}{dx}\frac {dx}{dt}=p_x'p$$ $\endgroup$ – Isham Nov 11 '18 at 17:44
  • $\begingroup$ The independent variable is $t$ not $x$. $\endgroup$ – Mohammad Riazi-Kermani Nov 11 '18 at 17:48
  • $\begingroup$ He's making a change of variable so op can integrate the equation..Mohammed thats the point And remember that t dosent appear in the equation $\endgroup$ – Isham Nov 11 '18 at 17:49
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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 $$ x = 1 + 2 \left(\sqrt{t+1}-1 \right) \exp \sqrt{t+1} $$

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