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$$x\left(\dfrac{dy}{dx}\right)^2 - \dfrac{dy}{dx} = x^4$$

I tried to integrate both sides w.r.t $x$, but it seems the resulting expressions are actually more complicated.

EDIT: Treating this as a quadratic in $dy/dx$, we have the discriminant $4x^5 + 1$. If we ask for rational solutions, there should exist a rational $m$ such that $4x^5 + 1 = m^2$. Hence, the question is now on the rational solutions to the equation $4x^5 + 1 = m^2$.

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    $\begingroup$ It's a quadratic equation: solve for $dy/dx$ using the classic formula. You will then get $dy/dx = g(x)$ for some function $x$ which can be integrated directly. $\endgroup$ – Winther Jan 10 '16 at 9:05
  • $\begingroup$ @Winther. The integration of $g(x)$ does not look very pleasant to me. Any idea on your side ? $\endgroup$ – Claude Leibovici Jan 10 '16 at 9:12
  • $\begingroup$ If you ask for rational solutions, shouldn't $m \geqslant 0$ ? $\endgroup$ – Airdish Jan 10 '16 at 9:15
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    $\begingroup$ @ClaudeLeibovici I think substituting $\sqrt{1+4x^5}$ should work. With $z= \sqrt{1+4x^5}$ the integrand becomes $\frac{\sqrt{1+4x^5}}{x}dx \to \frac{2 z^2}{5 \left(z^2-1\right)}dz$ which is easy to handle ($\text{\arctanh}$ solution). $\endgroup$ – Winther Jan 10 '16 at 9:19
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    $\begingroup$ @Ant If $y$ has degree $n$ then $x(dy/dx)^2 - (dy/dx)$ has degree $2n-1$ which cannot be equal to $4$. $\endgroup$ – Winther Jan 10 '16 at 9:26
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HINT:

$$x\left(\frac{\text{d}y}{\text{d}x}\right)^2-\frac{\text{d}y}{\text{d}x}=x^4\Longleftrightarrow$$ $$xy'(x)^2-y'(x)=x^4\Longleftrightarrow$$ $$xy'(x)^2-y'(x)-x^4=0\Longleftrightarrow$$


Using the 'quadratic formula':


$$y'(x)=\frac{-(-1)\pm\sqrt{(-1)^2-4\cdot x\cdot (-x^4)}}{2\cdot x}\Longleftrightarrow$$ $$y'(x)=\frac{1\pm\sqrt{1--4x(x^4)}}{2\cdot x}\Longleftrightarrow$$ $$y'(x)=\frac{1\pm\sqrt{1+4x^5}}{2\cdot x}\Longleftrightarrow$$ $$y'(x)=\frac{1\pm\sqrt{1+4x^5}}{2x}\Longleftrightarrow$$ $$\int y'(x)\space\text{d}x=\int\frac{1\pm\sqrt{1+4x^5}}{2x}\space\text{d}x\Longleftrightarrow$$ $$y(x)=\int\frac{1\pm\sqrt{1+4x^5}}{2x}\space\text{d}x$$


For the integrals, notice:

$$\int\frac{1+\sqrt{1+4x^5}}{2x}\space\text{d}x=\frac{1}{2}\int\left[\frac{\sqrt{1+4x^5}}{x}+\frac{1}{x}\right]\space\text{d}x=$$ $$\frac{1}{2}\left[\int\frac{\sqrt{1+4x^5}}{x}\space\text{d}x+\int\frac{1}{x}\space\text{d}x\right]=\frac{1}{2}\left[\int\frac{\sqrt{1+4x^5}}{x}\space\text{d}x+\ln|x|\right]$$


$$\int\frac{1-\sqrt{1+4x^5}}{2x}\space\text{d}x=\frac{1}{2}\int\left[\frac{1}{x}-\frac{\sqrt{1+4x^5}}{x}\right]\space\text{d}x=$$ $$\frac{1}{2}\left[\int\frac{1}{x}\space\text{d}x-\int\frac{\sqrt{1+4x^5}}{x}\space\text{d}x\right]=\frac{1}{2}\left[\ln|x|-\int\frac{\sqrt{1+4x^5}}{x}\space\text{d}x\right]$$

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  • $\begingroup$ @Isaac. Maybe there is, but evaluate the integral and look for them :) $\endgroup$ – Jan Jan 10 '16 at 13:50

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