1
$\begingroup$

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Page 633 in the foreign book I have been doing -- sorry no English version available and the book has not been peer-reviewed.

  • 0
    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 $T''(x) = - e^x$.2012-02-21

2 Answers 2

2

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.

  • 0
    I would stick to Stephen Montgomery-Smith's comment.2013-11-06
0

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