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I've started like this:

From $xy'^3 - 2yy'^2 - 16x^2 = 0$, we find that $$y = -8x^2/p^2 + 1/2 \cdot xp,$$ with $p = y' = dy/dx$.

Taking the derivative yields $$p = -16x/p^2 + 16 x^2p'/p^3 + 1/2p + 1/2 \cdot xp'$$ And working out the denominator gives us $$p^4 = -32xp + 32x^2 p' + xp^2p'.$$ At this point, I don't know how to go on. I tried separating the terms containing $p$ and $p'$, but it didn't get me anywhere.

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  • $\begingroup$ Maple solves this equation! $\endgroup$ – Dr. Sonnhard Graubner Aug 7 '17 at 19:55
  • $\begingroup$ What is a DFE? If you're going to use abbreviations that you invented yourself, you probably need to define them for us. $\endgroup$ – bof Aug 9 '17 at 8:05
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We are given

$$\tag 1 xy'^3 - 2yy'^2 - 16x^2 = 0$$

We will rewrite $(1)$ using $p = \dfrac{dy}{dx}$ as

$$\tag 2 x p^3 - 2 y p^2 - 16 x^2 = 0$$

Isolating $y$ in $(2)$, we have

$$\tag 3 y = \dfrac{1}{2}( x p - 16 x^2 p^{-2})$$

Differentiating $(3$), we have

$$\tag 4 y ' = p = \dfrac{1}{2}(p + x p' - 32~ x~ p^{-2} + 32 ~x^2~ p^{-3}p')$$

Factoring $(4)$, we have

$$\tag 5 \dfrac{(32 x + p^3)(p - x p')}{2 p^3} = 0$$

From $(5)$, we solve for the two expressions

$$p^3 + 32 x = 0 \\ p - x p' = 0$$

I will assume you can take it from here.

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  • $\begingroup$ Thanks for the detailed answer! One question though: in (4); why is $y' = 0$? $\endgroup$ – querty Aug 8 '17 at 15:18
  • $\begingroup$ One should not ignore the envelope $$y=-\frac12(x(32x)^{1/3}+16x^2(32x)^{-2/3})$$ of the solution family $$y=\frac12(Cx^2-16C^{-2}).$$ Piecewise combinations of these functions are also solutions. $\endgroup$ – LutzL Aug 8 '17 at 21:35
  • $\begingroup$ @LutzL: Good point - I didn't say that at first, but then changed my answer to ignore it - but I will change it back to my original solution. Thanks $\endgroup$ – Moo Aug 8 '17 at 21:36
  • $\begingroup$ I think you meant $y' = p$ instead of $y' = \frac{\mathrm{d}p}{\mathrm{d}x}$ in line 4. $\endgroup$ – Michael Lee Aug 8 '17 at 22:10
  • $\begingroup$ @MichaelLee: You are correct - typos! Corrected - thanks! $\endgroup$ – Moo Aug 8 '17 at 23:06
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$$p^4 = -32xp + 32x^2 p' + xp^2p'.$$ $$(32x^2 + xp^2)p'=p^4 +32xp .$$ This is an generalized Abel's equation of the second kind : $$y'\sum_{j=0}^{2}g_j(x)y^j=\sum_{j=0}^{4}f_j(x)y^j$$ where $g_0(x)=32x^2 \:;\:g_2(x)=x \:;\: f_1(x)=32x \:;\: f_4(x)=1 \:;\:g_1(x)=f_0(x)=f_2(x)=f_3(x)=0 .$

https://www.encyclopediaofmath.org/index.php/Abel_differential_equation

The majority of this kind of equations are not solvable on a closed form. Nevertheless, some particular solutions can sometimes be found. Knowing a boundary condition can eventually help.

In the case of the original equation : $$xy'^3 - 2yy'^2 - 16x^2 = 0$$ looking for a particular solution on the form $y_p(x)=ax^p$ leads to : $$y_p=-\frac{3}{\sqrt[3]{2}}x^{4/3}$$

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  • $\begingroup$ Notice that you can cancel the factor $32x+p^3$ in the second equation, as long as it is not zero. Then you are back to the solutions in the answer of Moo. -- By chance the monomial family indeed gives a solution which is the envelope of the parabolic solutions. $\endgroup$ – LutzL Aug 9 '17 at 10:01
  • $\begingroup$ @JJacquelin: Very nice observations regarding that this is a generalized Abel equation - I didn't even see that. (+1). $\endgroup$ – Moo Aug 9 '17 at 14:27

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