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In the following solution of the Korteweg-deVries PDE $$ u_t + 6uu_x + u_{xxx} = 0 \qquad (3.1) $$ I do not understand the second integration step and how they arrive at the expression for the differentials.

The first integration is clear to me, but in the second step, why $u_{\xi}$ does not become $u$ but $1/2 u_{\xi}^2$? And how they came from the equation $$ \frac{1}{2}u_{\xi}^2 = -u^3 + \frac{1}{2}cu^2 + c_2 $$ to $$ d\xi = \frac{du}{u\sqrt{c-2u}} $$ Btw I always find these calculations with differentials a little bit intimidating. enter image description here

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Before the second integration, the whole equation is multiplied by $u_\xi$ (a standard trick, but perhaps it should have been explained in the text), and then the chain rule is used backwards.

And if $c_2=0$, then $$ \frac12 \left( \frac{du}{d\xi} \right)^2 = -u^3 + \frac12 c u^2 = \frac12 u^2 (c-u) , $$ so $$ \frac{du}{d\xi} = \pm\sqrt{u^2 (c-u)} . $$ At this point, the author is somewhat unclear/lazy/sloppy when it comes to explaining why only the case $\frac{du}{d\xi} = u \sqrt{c-u}$ is considered, but if we accept that, then the rest is just the standard method for integrating a separable ODE. (The computation with differentials can be seen as just a mnemonic; it has been discussed in other questions on this site.)

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