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[tex]

\frac{{dy}}{{dx}}\frac{{d^3 y}}{{dx^3 }} - 2\left( {\frac{{d^2 y}}{{dx^2 }}} \right)^2 + \left( {\frac{{dy}}{{dx}}} \right)^2 = 0

[/tex]

Answer: cosech(Ay+B) +coth (Ay+B) = exp(-x).

I haven't gotten anywhere with this question but here are my thoughts. Normally when I see an equation with products of derivatives I try to 'reverse' the chain rule to see if I can write the LHS as a derivative with respect to the independent variable. However I think that method only works for linear DEs and I don't think this one is linear. Looking at this equation I'd probably substitute [tex]p = \frac{{dy}}{{dx}}[/tex]. This should give me:

[tex]

p\frac{{d^2 p}}{{dx^2 }} - 2\left( {\frac{{dp}}{{dx}}} \right)^2 + p^2 = 0

[/tex]

The book says make the same substitution as I did but the resulting equation seems to suggest that they've treated the equation for the case where the independent variable is not present in an explicit form. The way I've done it assumes that the dependent variable(in this case being y) is missing in an explicit form, which for this equation seems to be true. Anyway, after making the same substitution as I did, according to the book the resulting equation should be:

[tex]

p\frac{{d^2 p}}{{dy^2 }} + 1 = \left( {\frac{{dp}}{{dy}}} \right)^2

[/tex]

Basically I'm not really sure how to start this question. Can someone explain what kind of a substitution should be made, how substituted variables should be treated(if understanding of this particular aspect is necessary), and if possible, the reasoning behind the required substitution? Any help would be great, thanks.