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Page 633 in the foreign book I have been doing -- sorry no English version available and the book has not been peer-reviewed.

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Did you tried using Wolfram Alpha? It can also shows you a step-by-step solution. –  FUZxxl Feb 16 '12 at 20:16
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"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
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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

2 Answers 2

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.

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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

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

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

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