Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The last time when I thought that the task was about solving a non-linear differential equation with convolution and Frobenius -method more here, my instructor cheered me up that the goal was some sort of numerical approximation for the non-linear differential equation with first-order differentials -- it used the term to solve in "system form" which was at best misleading (well I never understood how it really was meant like that but here is a next puzzle). The task is to solve this "with integrals" or actually it uses some slang "kvadratures" in the assignment. I am now unsure whether I should use many times chain-rule and solve it with brute-force or whether there is some elegant way to solve this with "with kvadratures"? I am not sure whether the author is now referring to some numerical method or does it mean really just to integrate it and solve it?

Page 633 in the foreign book I have been doing -- sorry no English version available and the book has not been peer-reviewed.

share|improve this question
Did you tried using Wolfram Alpha? It can also shows you a step-by-step solution. –  FUZxxl Feb 16 '12 at 20:16
"kvadratures" in English is "quadratures", meaning integrals: a solution involving integrals (which may or may not be possible to do in closed form). –  Robert Israel Feb 16 '12 at 20:34
In this case "quadratures" is a bit of a red herring. Hint: change of variables $y(t) = x, t = T(x)$. –  Robert Israel Feb 16 '12 at 20:43
@RobertIsrael: $y''=(y(t))''=(x)''=(1)'=0$ and $(y(t)')^{3}=(x')^{3}=1^{3}=1$ so $0=e^{y}$ and this is absurd?! $y(t)=x$? But why? And now some $t=T(x)$ but for which reason? –  hhh Feb 16 '12 at 21:02
Let $v = \frac{dy}{dt} = \frac{1}{T'(x)}$. Then $\frac{d^2y}{dt^2} = \frac{dv}{dt} = \frac{1}{T'(x)} \frac{dv}{dx} = -\frac{T''(x)}{T'(x)^3}$, and the differential equation becomes $-\frac{T''(x)}{T'(x)^3} = \frac{1}{T'(x)^3} e^x$ or $T''(x) = - e^x$. –  Robert Israel Feb 21 '12 at 3:10
add comment

2 Answers

up vote 2 down vote accepted

You have

$$y'' = (y')^3 e^{y}$$

$$\dfrac{y''}{y'^2} = y' e^{y}$$

$$-d\left\{\dfrac{1}{y'}\right\} = d\{ e^{y}\}$$

$$-\dfrac{1}{y'}= e^y+C$$

$$-1= y' e^y+Cy'$$

$$C_1-x= e^y+Cy$$

See this question here on how to use the Lambert W -function to solve this problem.

share|improve this answer
Hello Pedro, I am not sure but is the Lambert W function related to this question here? If I understand correctly, the generating function of Abel's polynomial has a Lambert W function and the integral has Abel's polynomial. Able to look at the integral? –  hhh Nov 6 '13 at 19:28
I would stick to Stephen Montgomery-Smith's comment. –  Pedro Tamaroff Nov 6 '13 at 19:47
add comment

substitute : $y'_x= v \Rightarrow y''_x=v'_y\cdot v~$ so :

$$v'_y \cdot v = v^3 \cdot e^y \Rightarrow v'_y =v^2 \cdot e^y \Rightarrow \frac {dv}{dy}=v^2 \cdot e^y \Rightarrow \int \frac{dv}{v^2}=\int e^y \,dy$$

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.