# How can this differential equation be solved?

$$2x''\ln (x') =x' \,\, x(0)=1, x'(0)=e$$ 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

• I hope I didn't made mistakes .. – Isham Nov 11 '18 at 17:24
• 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 $p(t)$. – Alejandro Nasif Salum Nov 11 '18 at 17:30
• @AlejandroNasifSalum p' is according to x not t – Isham Nov 11 '18 at 17:39
• You mean $p'=\frac{dp}{dx}$? – Alejandro Nasif Salum Nov 11 '18 at 17:43
• @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 $p$. – LutzL Nov 11 '18 at 18:29

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 ?

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 $$x(0)=1$$.

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.

• Thats correct Mohammaed the derivative p' is according to x not t so we have $p'p=x''$ – Isham Nov 11 '18 at 17:38
• @Isham Are you sure? Think about it again. – Mohammad Riazi-Kermani Nov 11 '18 at 17:43
• Mohammed try it yourself $$x''=\frac {dp}{dt}=\frac {dp}{dx}\frac {dx}{dt}=p_x'p$$ – Isham Nov 11 '18 at 17:44
• The independent variable is $t$ not $x$. – Mohammad Riazi-Kermani Nov 11 '18 at 17:48
• 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 – Isham Nov 11 '18 at 17:49

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 $$x = 1 + 2 \left(\sqrt{t+1}-1 \right) \exp \sqrt{t+1}$$