Is there an analytic solution to the following equation?
$$ \frac{dx}{dy} = \frac{x^2 - y^2}{x^2 + y^2} $$
I believe the answer is 'no'. It isn't separable or exact I've had trouble finding any documentation that can help me explore this problem further. I have solved the problem numerically to explore the solution space, and it seems to make sense as the answer, but I also need to show some work in attempting to solve it analytically.
Is there a substitution that can be done to solve this? Or is this, in fact, an unsolvable problem (analytically)?
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$\begingroup$The following procedure gives more geometrical insight, but in the end leads to cumbersome calculations as well. Put $$x:=r(\phi)\cos\phi,\quad y:=r(\phi)\sin\phi$$ with the polar angle $\phi$ as new independent variable and unknown function $\phi\mapsto r(\phi)$. Then $$x'=r'\cos\phi-r\sin\phi,\qquad y'=r'\sin\phi+r\cos\phi\ .$$ In this way we obtain $${r'\cos\phi-r\sin\phi\over r'\sin\phi+r\cos\phi}={dx\over dy}={x^2-y^2\over x^2+y^2}=\cos(2\phi)\ ,$$ or $${r'\over r}={\sin\phi+\cos\phi\cos(2\phi)\over\cos\phi-\sin\phi\cos(2\phi)}=:f(\phi)\ .$$ Determine a primitive $F$ of $f$ in a maximal interval containing $\phi=0$, and you obtain a global description of the solution curves.
$\endgroup$ $\begingroup$Given homogenius ODE is$$\dfrac{\mathrm dy}{\mathrm dx} = -\dfrac{y^2+x^2}{y^2-x^2}.$$Can be applied substitution$$y= (u-1)x,\tag1$$$$x\dfrac{\mathrm du}{\mathrm dx} = -u+1-\dfrac{(u-1)^2+1}{(u-1)^2-1},$$$$x\dfrac{\mathrm du}{\mathrm dx}=-u-\dfrac{2}{(u-1)^2-1},$$
$$\dfrac{\mathrm dx}{x}=\dfrac{2u-u^2}{u^3-2u^2+2}{\mathrm du}.\tag2$$Since the denominator of RHS(2) has a root$$r=\dfrac13(2-\sqrt[3]{19+\sqrt{297}}-\sqrt[3]{19-\sqrt{297}})\approx-0.83928\,67552\,14161,\tag3$$\begin{align} &27(r^3-2r^2+2) = (3r-2)^3 - 12(3r-2) + 38\\ &= -\left(\sqrt[3]{19+\sqrt{297}\,}\,+\sqrt[3]{19-\sqrt{297}\,}\,\right)^3 +12\left(\sqrt[3]{19+\sqrt{297}}+\sqrt[3]{19-\sqrt{297}}\right)+38\\ &= \left(-3\sqrt[3]{19+\sqrt{297}\,}\,\sqrt[3]{19-\sqrt{297}\,}+12\right)\left(\sqrt[3]{19+\sqrt{297}\,}\,+\sqrt[3]{19-\sqrt{297}\,}\,\right)\\ &= \left(-3\sqrt[3]{64}+12\right)\left(\sqrt[3]{19+\sqrt{297}\,}\,+\sqrt[3]{19-\sqrt{297}\,}\,\right) = 0, \end{align}then$$u^3-2u^2+2 = (u-r)(u^2-(2-r)u+r^2-2r),$$$$R(u)=\dfrac{2u-u^2}{u^3-2u^2+2}=\dfrac A{u-r}+\dfrac{B(2u-2+r)}{u^2-(2-r)u+r^2-2r} +\dfrac{C}{u^2-(2-r)u+r^2-2r},$$where$$A = \lim_{u\to r}(u-r)R(u) = \dfrac{2r-r^2}{3r^2-4r} = -\dfrac{2-r}{4-3r},$$$$A+2B = \lim_{u\to \infty}uR(u) = -1,\quad B=-\dfrac{1-r}{4-3r},$$$$\dfrac {B-A}r+\dfrac C{r^2-2r} = R(0)=0,\quad C=\dfrac{2-r}{4-3r},$$$$u^2-(2-r)u+r^2-2r = \left(u-\dfrac{2-r}2\right)^2+\dfrac{(2-r)(-3r-2)}4.$$
Then from $(2)$ should$$(4-3r)\ln \mathrm{const}|x| = -(2-r)\ln|u-r|-(1-r)\ln|u^2-(2-r)u+r^2-2r|\\ +2\sqrt{\dfrac{2-r}{-3r-2}}\arctan\dfrac{2u-2+r}{\sqrt{(2-r)(-3r-2)}},$$$$\color{green}{\mathbf{\small C_1|x|^{4-3r}|u-r||u^3-2u^2-2|^{1-r} = \exp\left(2\sqrt{\dfrac{2-r}{-3r-2}}\arctan\dfrac{2u-2+r}{\sqrt{(2-r)(-3r-2)}}\right)}},\tag4$$where $C_1$ is the arbitrary constant.
Formulas $(1),(3),(4)$ present the common solution of the given ODE.
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